🔬 Model Lab

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** Find the sum of the two smallest prime factors of 250.

**Solution:**
First, I'll find the prime factorization of 250.
- 250 = 2 × 125
- 250 = 2 × 5 × 25
- 250 = 2 × 5 × 5 × 5
- 250 = 2 × 5³

The prime factors of 250 are: 2 and 5

The two smallest prime factors are 2 and 5.
Sum = 2 + 5 = 7

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How many different four-digit numbers can be formed by rearranging the digits in 2004?

**Solution:**
The digits are: 2, 0, 0, 4
We have 4 digits total, but two of them are identical (two 0's).

If all digits were distinct, we'd have 4! = 24 arrangements.
Since we have two identical 0's, we divide by 2!:
Number of arrangements = 4! / 2! = 24 / 2 = 12

However, we must exclude arrangements that start with 0 (these aren't four-digit numbers).
- Arrangements starting with 0: We arrange the remaining 3 digits (2, 0, 4) = 3! / 1! = 6

Valid four-digit numbers = 12 - 6 = 6

ANSWER 2: B

---

**Problem 3:**

**What is being asked:** If m and n are positive odd integers, which expression must also be odd?

**Solution:**
Let me test each option. Let m = 1, n = 3 (both odd).

A. m + 3n = 1 + 9 = 10 (even) ✗

B. 3m − n = 3(1) − 3 = 0 (even) ✗

C. 3m² + 3n² = 3(1)² + 3(3)² = 3 + 27 = 30 (even) ✗

D. (nm + 3)² = (1·3 + 3)² = (6)² = 36 (even) ✗

E. 3mn = 3(1)(3) = 9 (odd) ✓

Let me verify E generally: m and n are odd, so mn is odd (odd × odd = odd).
Then 3mn = 3 × odd = odd ✓

ANSWER 3: E

---

**Problem 4:**

**What is being asked:** In a jar with red, green, and blue marbles, all but 6 are red, all but 8 are green, all but 4 are blue. How many marbles total?

**Solution:**
Let R = red, G = green, B = blue, T = total marbles.

- "All but 6 are red" means: G + B = 6
- "All but 8 are green" means: R + B = 8
- "All but 4 are blue" means: R + G = 4

Adding all three equations:
2R + 2G + 2B = 18
R + G + B = 9

Total marbles = 9

ANSWER 4: C

---

**Problem 5:**

**What is being asked:** Which number is the largest?

**Solution:**
Comparing by looking at each decimal place:
- A. 0.97
- B. 0.979
- C. 0.9709
- D. 0.907
- E. 0.9089

All start with 0.9. Looking at the second decimal place: A and B have 7, C has 7, D has 0, E has 0.
So D and E are smaller.

Between A, B, and C:
- A. 0.97 = 0.9700
- B. 0.979 = 0.9790
- C. 0.9709 = 0.9709

Comparing: 0.9700 vs 0.9790 vs 0.9709
At the third decimal: 0 vs 9 vs 0
B has 9 in the third position, so B = 0.979 is largest.

ANSWER 5: B

---

**Problem 6:**

**What is being asked:** By how many cm does water level rise when a 1000 cm³ rock is submerged?

**Solution:**
Base area = 100 cm × 40 cm = 4000 cm²

When the rock is submerged, it displaces water equal to its volume.
The water level rise = Volume / Base Area = 1000 / 4000 = 0.25 cm

ANSWER 6: A

---

**Problem 7:**

**What is being asked:** Maximum number of principals during an 8-year period if each serves exactly one 3-year term?

**Solution:**
Each principal serves 3 years. In an 8-year period, we want maximum principals.

Timeline strategy:
- Principal 1: Years 0-3 (ends at year 3)
- Principal 2: Years 3-6 (starts at year 3)
- Principal 3: Years 6-9 (starts at year 6)

Within an 8-year period (years 0-8), Principals 1, 2, and 3 all serve.
We could also have a principal starting at year 6 ending at year 9, but only years 6-8 are in our window.

Actually, maximum: Start at year 0, have principals at years 0-3, 3-6, 6-8 (partial).
That's 3 full principals minimum. Could we fit a 4th?

If we start right before year 0: years -1 to 2, then 2-5, then 5-8. That's 3 principals fully within the 8-year window.

Maximum is 3 principals.

ANSWER 7: B

---

**Problem 8:**

**What is being asked:** How many hours must Theresa work in week 6 to average 10 hours/week over 6 weeks?

**Solution:**
Total hours needed = 10 hours/week × 6 weeks = 60 hours

Hours in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48 hours

Hours needed in week 6 = 60 - 48 = 12 hours

ANSWER 8: D

---

**Problem 9:**

**What is being asked:** Find the missing digit A in 1A2, where 11 members each paid the same integer amount totaling 1A2 dollars.

**Solution:**
The three-digit number is 1A2, where A is a digit (0-9).
This represents: 100 + 10A + 2 = 102 + 10A

This must be divisible by 11 (since 11 members each paid the same integer amount).

Testing values:
- A = 0: 102 ÷ 11 = 9.27... ✗
- A = 1: 112 ÷ 11 = 10.18... ✗
- A = 2: 122 ÷ 11 = 11.09... ✗
- A = 3: 132 ÷ 11 = 12 ✓
- A = 4: 142 ÷ 11 = 12.9... ✗

So A = 3, and 132 ÷ 11 = 12 dollars per person.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** Average rainfall per hour for 366 inches in July 1861.

**Solution:**
July has 31 days.
Hours in July = 31 days × 24 hours/day

Average rainfall per hour = Total inches / Total hours
= 366 inches / (31 × 24 hours)

ANSWER 10: A

---

**Problem 11:**

**What is being asked:** How many bananas did Big Al eat on May 5?

**Solution:**
Each day he ate 6 more than the previous day. Let day 1 (May 1) = x bananas.
- May 1: x
- May 2: x + 6
- May 3: x + 12
- May 4: x + 18
- May 5: x + 24

Total = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60 = 100

5x = 40
x = 8

May 5 = x + 24 = 8 + 24 = 32 bananas

ANSWER 11: D

---

**Problem 12:**

**What is being asked:** Combined average age of two groups (6 people at age 40 average, 4 people at age 25 average).

**Solution:**
Total age in Room A = 6 × 40 = 240 years
Total age in Room B = 4 × 25 = 100 years

Combined total age = 240 + 100 = 340 years
Combined total people = 6 + 4 = 10 people

Average age = 340 / 10 = 34 years

ANSWER 12: D

Problem 1:

We want the sum of the two smallest prime factors of 250.

First factor 250:
- \(250 = 2 \times 125\)
- \(125 = 5 \times 25 = 5^3\)

So the prime factorization is:
\[
250 = 2 \cdot 5^3
\]

The two smallest prime factors are:
- \(2\)
- \(5\)

Their sum is:
\[
2+5=7
\]

So the correct choice is C.

ANSWER 1: C

Problem 2:

We are rearranging the digits in 2004 to make different four-digit numbers.

The digits are: \(2,0,0,4\)

A four-digit number cannot start with 0, so we count distinct arrangements of these digits that begin with 2 or 4.

### Case 1: First digit is 2
Then the remaining digits are \(0,0,4\).  
The number of distinct rearrangements of these 3 digits is:
\[
\frac{3!}{2!}=3
\]

### Case 2: First digit is 4
Then the remaining digits are \(0,0,2\).  
Again, the number of distinct rearrangements is:
\[
\frac{3!}{2!}=3
\]

Total:
\[
3+3=6
\]

So the correct choice is B.

ANSWER 2: B

Problem 3:

We are told \(m\) and \(n\) are positive odd integers. We need the expression that must be odd.

Recall:
- odd + odd = even
- odd \(\times\) odd = odd
- odd + even = odd
- even squared is even, odd squared is odd

Check each choice.

### A. \(m + 3n\)
- \(n\) is odd, so \(3n\) is odd.
- \(m\) is odd.
- odd + odd = even.

So A is even, not odd.

### B. \(3m - n\)
- \(3m\) is odd.
- odd - odd = even.

So B is even.

### C. \(3m^2 + 3n^2\)
- \(m^2\) is odd, so \(3m^2\) is odd.
- \(n^2\) is odd, so \(3n^2\) is odd.
- odd + odd = even.

So C is even.

### D. \((nm + 3)^2\)
- \(nm\) is odd \(\times\) odd = odd.
- odd + 3 = odd + odd = even.
- even squared = even.

So D is even.

### E. \(3mn\)
- \(mn\) is odd \(\times\) odd = odd.
- \(3mn\) = odd \(\times\) odd = odd.

So E must be odd.

Therefore the correct choice is E.

ANSWER 3: E

Problem 4:

We are told:
- all but 6 are red
- all but 8 are green
- all but 4 are blue

Let the total number of marbles be \(N\).

Then:
- red marbles = \(N - 6\)
- green marbles = \(N - 8\)
- blue marbles = \(N - 4\)

Adding these gives:
\[
(N-6) + (N-8) + (N-4) = 3N - 18
\]

But this sum also counts every marble exactly once, because every marble is either red, green, or blue. So:
\[
3N - 18 = N
\]

Solve:
\[
2N = 18
\]
\[
N = 9
\]

So the correct choice is C.

ANSWER 4: C

Problem 5:

We compare the decimals:

- A: 0.97 = 0.9700
- B: 0.979
- C: 0.9709
- D: 0.907
- E: 0.9089

Now compare:
- \(0.979\) is larger than \(0.9709\) and \(0.9700\)
- \(0.9089\) and \(0.907\) are both smaller than 0.97

So the largest is \(0.979\).

Correct choice: B.

ANSWER 5: B

Problem 6:

The aquarium has base \(100 \text{ cm} \times 40 \text{ cm}\), so its base area is:
\[
100 \cdot 40 = 4000 \text{ cm}^2
\]

A rock of volume \(1000 \text{ cm}^3\) is completely submerged, so it displaces \(1000 \text{ cm}^3\) of water.

The rise in water level is:
\[
\text{rise} = \frac{\text{volume displaced}}{\text{base area}} = \frac{1000}{4000} = 0.25 \text{ cm}
\]

So the correct choice is A.

ANSWER 6: A

Problem 7:

Each principal serves exactly one 3-year term. We want the maximum number of principals during an 8-year period.

To maximize the number of different principals, we want each principal to serve without overlap as much as possible.

In 8 years, we can fit:
- one principal for years 1–3
- another for years 4–6
- another for years 7–8, but a principal must serve a full 3-year term

So the maximum number of full 3-year terms that can begin within an 8-year period is:
\[
\left\lceil \frac{8}{3} \right\rceil = 3
\]

But let's check carefully: if a principal starts in year 1, the term runs through year 3. Another can start in year 4 and run through year 6. Another can start in year 7 and run through year 9, and the school still has a principal during the 8-year period. Thus 3 principals is possible.

Could there be 4 principals? No, because 4 full 3-year terms would require 12 years if non-overlapping.

So the maximum is 3.

Correct choice: B.

ANSWER 7: B

Problem 8:

Theresa must average 10 hours per week for 6 weeks, so total hours needed are:
\[
10 \times 6 = 60
\]

For the first 5 weeks, she worked:
\[
8 + 11 + 7 + 12 + 10 = 48
\]

Let \(x\) be the number of hours in the 6th week:
\[
48 + x = 60
\]
\[
x = 12
\]

So she must work 12 hours.

Correct choice: D.

ANSWER 8: D

Problem 9:

The total paid is written as \(\$1A2\), a 3-digit number, and 11 members paid the same integer amount.

So the total must be divisible by 11.

We need the digit \(A\) so that \(1A2\) is divisible by 11.

For a 3-digit number \(abc\), divisibility by 11 means:
\[
a - b + c \equiv 0 \pmod{11}
\]

Here:
\[
1 - A + 2 = 3 - A
\]

This must be a multiple of 11. Since \(A\) is a digit 0–9, \(3-A\) must be 0 or \(\pm 11\).

The only digit that works is:
\[
3 - A = 0 \Rightarrow A = 3
\]

Check:
\[
132 \div 11 = 12
\]
So it works.

Correct choice: D.

ANSWER 9: D

Problem 10:

We are asked for average rainfall in inches per hour during July 1861.

July has 31 days, and each day has 24 hours, so the total number of hours is:
\[
31 \times 24
\]

Average rainfall per hour is:
\[
\frac{366}{31 \times 24}
\]

That matches choice A.

ANSWER 10: A

Problem 11:

Big Al ate 100 bananas in 5 days, with each day 6 more than the previous day.

Let the number eaten on May 1 be \(x\). Then the amounts are:
\[
x,\ x+6,\ x+12,\ x+18,\ x+24
\]

Their sum is 100:
\[
x + (x+6) + (x+12) + (x+18) + (x+24) = 100
\]
\[
5x + 60 = 100
\]
\[
5x = 40
\]
\[
x = 8
\]

Then on May 5:
\[
x+24 = 8+24 = 32
\]

So the correct choice is D.

ANSWER 11: D

Problem 12:

Room A:
- 6 people
- average age 40
So total age in Room A:
\[
6 \times 40 = 240
\]

Room B:
- 4 people
- average age 25
So total age in Room B:
\[
4 \times 25 = 100
\]

Combined:
- total people = \(6+4=10\)
- total age = \(240+100=340\)

Average age:
\[
\frac{340}{10} = 34
\]

So the correct choice is D.

ANSWER 12: D

Problem 1:  
We need the sum of the two smallest prime factors of 250.  
Factor 250:
\[
250=25\cdot 10=5^2\cdot (2\cdot 5)=2\cdot 5^3
\]
Prime factors are \(2, 5, 5, 5\). The two smallest prime factors (counting multiplicity) are \(2\) and \(5\).  
Sum:
\[
2+5=7
\]
Tempting wrong choices:  
- \(10\) would be \(2+8\) (not a prime factor).  
- \(12\) would require primes like \(2+10\) etc., but only \(2\) and \(5\) appear.  
ANSWER 1: C

---

Problem 2:  
How many different four-digit numbers can be formed by rearranging the four digits in 2004?  
Digits: \(2,0,0,4\). There are 4 positions, but two digits are repeated (two 0’s).  
Total distinct arrangements:
\[
\frac{4!}{2!}= \frac{24}{2}=12
\]
However, this counts arrangements where the first digit could be 0, which would not be a four-digit number.  
Count four-digit arrangements by fixing the first digit:

- First digit cannot be 0, so it’s either 2 or 4.
- Case 1: First digit is 2. Remaining digits: \(0,0,4\) → number of distinct permutations:
\[
\frac{3!}{2!}=3
\]
- Case 2: First digit is 4. Remaining digits: \(0,0,2\) → again:
\[
\frac{3!}{2!}=3
\]
Total:
\[
3+3=6
\]
ANSWER 2: B

---

Problem 3:  
\(m\) and \(n\) are positive odd integers. Which expression must also be odd?  
Key fact: odd + odd = even, odd − odd = even, odd × odd = odd, odd\(^2\)=odd.

Check each choice:

A. \(m+3n\)  
\(3n\) is odd (odd×odd). odd + odd = even → not must be odd.

B. \(3m-n\)  
\(3m\) is odd. odd − odd = even → not.

C. \(3m^2+3n^2\)  
\(m^2\) is odd, so \(3m^2\) is odd; similarly \(3n^2\) is odd.  
odd + odd = even → not.

D. \((nm+3)^2\)  
\(nm\) = odd×odd = odd. odd + 3 (odd) = even.  
Even squared = even → not odd.

E. \(3mn\)  
\(mn\) is odd×odd = odd. \(3mn\) = odd×odd = odd.  
This must be odd.

ANSWER 3: E

---

Problem 4:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- all but 8 are green → green marbles = 8
- all but 4 are blue → blue marbles = 4

Total marbles:
\[
6+8+4=18
\]
But the choices are 6, 8, 9, 10, 12—so interpret carefully:

“All but 6 are red” means **all except 6 are red**, so red = total − 6.  
Similarly:
- green = total − 8  
- blue = total − 4  

Let total be \(T\). Then:
\[
\text{Red}=T-6,\quad \text{Green}=T-8,\quad \text{Blue}=T-4
\]
These sum to total:
\[
(T-6)+(T-8)+(T-4)=T
\]
\[
3T-18=T
\]
\[
2T=18 \Rightarrow T=9
\]
So there are 9 marbles total.  
ANSWER 4: C

---

Problem 5:  
Find the largest of:
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  

Compare decimals by extending with zeros if helpful:

- \(0.97 = 0.9700\)
- \(0.979 = 0.9790\)
- \(0.9709 = 0.9709\)

Now compare the ones with prefix 0.97:
- \(0.9709 > 0.9700\)

And compare \(0.9790\) with \(0.9709\):
\[
0.979 > 0.9709 \quad (\text{since } 0.979 \text{ has a larger thousandths part})
\]
So the largest is \(0.979\).  
ANSWER 5: B

---

Problem 6:  
Rectangular base: \(100 \text{ cm} \times 40 \text{ cm}\).  
Water depth initially: 37 cm, so initial water volume:
\[
V_i = (100\cdot 40)\cdot 37 = 4000\cdot 37 = 148{,}000\text{ cm}^3
\]
Rock volume placed and submerged: \(1000\text{ cm}^3\).  
Rock displaces water equal to its volume, so water volume increases by \(1000\text{ cm}^3\).  
Water surface rises by \(\Delta h\) where:
\[
(100\cdot 40)\Delta h = 1000
\]
\[
4000\Delta h = 1000
\]
\[
\Delta h = \frac{1000}{4000}=0.25
\]
ANSWER 6: A

---

Problem 7:  
Each principal serves exactly one 3-year term.  
During an 8-year period, how many different principals could serve?  
To maximize the number, start a new principal as soon as the previous term ends.

If one principal serves years \(1\)–\(3\), next serves \(4\)–\(6\), next serves \(7\)–\(9\).  
But we only have up to year 8, so the third principal would still be in office.

Count starts within 8 years:  
Possible term coverage:
- Principal 1: years 0–3 (or 1–3)
- Principal 2: next 3-year block
- Principal 3: next 3-year block

Essentially, number of full 3-year terms that can begin in 8 years:
\[
\text{Starts at year }1,\ 4,\ 7 \Rightarrow 3 \text{ principals}
\]
Can we have 4? That would require starts at 1, 4, 7, 10, but 10 is outside the 8-year window. So max is 3.

ANSWER 7: B

---

Problem 8:  
Average 10 hours per week for 6 weeks means total hours:
\[
10 \times 6 = 60
\]
Given first 5 weeks: \(8,11,7,12,10\). Sum:
\[
8+11+7+12+10=48
\]
Let final week be \(x\). Then:
\[
48+x=60 \Rightarrow x=12
\]
ANSWER 8: D

---

Problem 9:  
Eleven members each paid the same integer amount. Total paid:
\[
11 \cdot (\text{integer}) = 1A2
\]
where \(1A2\) is a 3-digit number with hundreds digit 1, tens digit \(A\), ones digit 2.

So:
\[
1A2 = 100 + 10A + 2 = 102 + 10A
\]
Must be divisible by 11. Check values:
\[
102, 112, 122, 132, 142 \quad \text{for } A=0,1,2,3,4
\]
Compute divisibility by 11:
- \(102/11 = 9.272...\) not
- \(112/11 = 10.181...\) not
- \(122/11 = 11.09...\) not
- \(132/11 = 12\) exactly (since \(11\cdot 12=132\))
- \(142/11\) not

So \(A=3\).  
ANSWER 9: D

---

Problem 10:  
Average rainfall per hour in July: July has 31 days.  
Hours in July:
\[
31 \times 24 = 744
\]
Total inches: 366.  
Average per hour:
\[
\frac{366}{744}
\]
Simplify by dividing top and bottom by 6:
\[
\frac{366}{744}=\frac{61}{124}
\]
But we just need the correct expression choice. The expression \(\frac{366}{31\times 24}\) matches choice A.  
ANSWER 10: A

---

Problem 11:  
Big Al eats 100 bananas from May 1 through May 5, increasing by 6 each day.  
Let May 1 be \(a\). Then:
\[
a,\ a+6,\ a+12,\ a+18,\ a+24
\]
Sum equals 100:
\[
5a + (0+6+12+18+24)=100
\]
Add the constants:
\[
0+6+12+18+24=60
\]
So:
\[
5a+60=100 \Rightarrow 5a=40 \Rightarrow a=8
\]
May 5 is:
\[
a+24=8+24=32
\]
ANSWER 11: D

---

Problem 12:  
Average age in Room A: 6 people average 40 → total age:
\[
6\cdot 40=240
\]
Room B: 4 people average 25 → total age:
\[
4\cdot 25=100
\]
Combined total people: \(6+4=10\). Combined total age:
\[
240+100=340
\]
Combined average:
\[
\frac{340}{10}=34
\]
ANSWER 12: D

Problem 1:
To find the prime factors of 250, we perform prime factorization: $250 = 25 \times 10 = (5^2) \times (2 \times 5) = 2 \times 5^3$. The prime factors are 2 and 5. The sum of these two smallest prime factors is $2 + 5 = 7$.
ANSWER 1: C

Problem 2:
The digits are $\{2, 0, 0, 4\}$. A four-digit number cannot start with 0. 
- If the first digit is 2: Remaining digits are $\{0, 0, 4\}$. Arrangements: $400, 400$ (Wait, the digits are 0, 0, 4. Permutations of 0, 0, 4 are $004, 040, 400$. So: 2004, 2040, 2400). That's 3 numbers.
- If the first digit is 4: Remaining digits are $\{2, 0, 0\}$. Permutations of 2, 0, 0 are $200, 020, 002$. So: 4200, 4020, 4002. That's 3 numbers.
Total = $3 + 3 = 6$.
ANSWER 2: B

Problem 3:
Let $m, n$ be odd. $m = 2k+1, n = 2j+1$.
A) $m+3n = (2k+1) + 3(2j+1) = 2k+1+6j+3 = 2(k+3j+2)$, which is even.
B) $3m-n = 3(2k+1) - (2j+1) = 6k+3-2j-1 = 6k-2j+2$, which is even.
C) $3m^2+3n^2 = 3(odd)^2 + 3(odd)^2 = 3(odd) + 3(odd) = odd + odd = even$.
D) $(nm+3)^2 = (odd \cdot odd + 3)^2 = (odd + 3)^2 = (even)^2 = even$.
E) $3mn = 3(odd)(odd) = 3(odd) = odd$.
ANSWER 3: E

Problem 4:
Let $R, G, B$ be the number of red, green, and blue marbles. Total $T = R+G+B$.
"All but 6 are red" $\implies G+B = 6$.
"All but 8 are green" $\implies R+B = 8$.
"All but 4 are blue" $\implies R+G = 4$.
Summing these: $2(R+G+B) = 6+8+4 = 18 \implies R+G+B = 9$.
ANSWER 4: C

Problem 5:
Compare digits place by place:
A) 0.9700
B) 0.9790
C) 0.9709
D) 0.9070
E) 0.9089
Comparing the thousandths place: B has 9, A/C have 7, D/E have 0. 0.979 is the largest.
ANSWER 5: B

Problem 6:
The volume of the water added is equal to the volume of the rock, $1000 \text{ cm}^3$. The base area of the aquarium is $100 \times 40 = 4000 \text{ cm}^2$. The rise in water level $h$ satisfies $4000 \times h = 1000$. Thus $h = 1000/4000 = 0.25$ cm.
ANSWER 6: A

Problem 7:
In 8 years, if one principal starts at time 0, they finish at year 3. The next starts at year 3 and finishes at year 6. The third starts at year 6 and finishes at year 9. Thus, 3 principals cover the 8-year period.
ANSWER 7: B

Problem 8:
Total hours needed for 6 weeks = $10 \times 6 = 60$.
Hours worked so far = $8+11+7+12+10 = 48$.
Hours needed in week 6 = $60 - 48 = 12$.
ANSWER 8: D

Problem 9:
The number $1A2$ is divisible by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. $1 - A + 2 = 3 - A$. For this to be divisible by 11, $3-A = 0 \implies A=3$.
ANSWER 9: D

Problem 10:
Total rainfall = 366 inches. Total hours = (days in July) $\times$ (hours per day) = $31 \times 24$. Average = Total / Total hours = $366 / (31 \times 24)$.
ANSWER 10: A

Problem 11:
Let $x$ be the bananas eaten on May 1.
Days: $x, x+6, x+12, x+18, x+24$.
Sum: $5x + 60 = 100 \implies 5x = 40 \implies x = 8$.
May 5: $x + 24 = 8 + 24 = 32$.
ANSWER 11: D

Problem 12:
Total age = (Sum in A) + (Sum in B) = $(6 \times 40) + (4 \times 25) = 240 + 100 = 340$.
Total people = $6 + 4 = 10$.
Average = $340 / 10 = 34$.
ANSWER 12: D

Problem 1:
The prime factorization of 250 is found by dividing successively by the smallest primes: \(250 = 2 \times 125 = 2 \times 5^3\). The distinct prime factors are therefore exactly 2 and 5. Their sum is \(2 + 5 = 7\). Among the choices, this matches C; the other options are either a single factor or sums that include larger or repeated factors not present in the factorization.

ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. The total number of distinct permutations of these four digits is \(4!/2! = 12\), but any arrangement with leading zero is not a four-digit number and must be excluded. The only admissible leading digits are therefore 2 or 4.
- Leading digit 2 leaves the multiset {0,0,4}; the distinct arrangements are 2004, 2040, 2400 (three numbers).
- Leading digit 4 leaves the multiset {0,0,2}; the distinct arrangements are 4002, 4020, 4200 (three numbers).
The total is therefore 6, matching choice B.

ANSWER 2: B

Problem 3:
Let \(m\) and \(n\) be positive odd integers. Each option is examined for parity:
- A: \(m + 3n\) is odd + odd = even.
- B: \(3m - n\) is odd - odd = even.
- C: \(m^2\) and \(n^2\) are both odd, so their sum is even and multiplication by 3 yields an even result.
- D: \(mn\) is odd, so \(mn + 3\) is even and its square is even.
- E: \(3mn\) is odd \(\times\) odd \(\times\) odd = odd.
Only E is necessarily odd.

ANSWER 3: E

Problem 4:
Let \(T\) be the total number of marbles. Then the three given conditions become the equations
\[
T - R = 6, \quad T - G = 8, \quad T - B = 4.
\]
Solving for each color gives \(R = T-6\), \(G = T-8\), \(B = T-4\). Substituting into \(T = R + G + B\) produces
\[
T = (T-6) + (T-8) + (T-4) \implies 18 = 2T \implies T = 9.
\]
Verification: 3 red, 1 green, 5 blue satisfies all three “all-but” statements. This value matches choice C.

ANSWER 4: C

Problem 5:
Aligning the decimals shows the values are
\[
0.9700,\quad 0.9790,\quad 0.9709,\quad 0.9070,\quad 0.9089.
\]
The largest entry is 0.9790, which is choice B.

ANSWER 5: B

Problem 6:
The base area of the aquarium is \(100 \times 40 = 4000\) cm\(^2\). The rock displaces a volume of 1000 cm\(^3\), so the rise in water level is the quotient
\[
\frac{1000}{4000} = 0.25
\]
cm. This matches choice A.

ANSWER 6: A

Problem 7:
Each principal occupies a non-overlapping 3-year term. In any contiguous 8-year window the maximum number of distinct 3-year blocks that can intersect the window is four. An explicit schedule realizing four principals is the blocks ending in years 1, 4, 7 and 10 (only the portions lying inside the 8-year interval are counted). Five principals would require at least nine years of coverage inside the window, which exceeds eight. Hence the maximum is 4, choice C.

ANSWER 7: C

Problem 8:
Six weeks at an average of 10 hours per week requires a total of 60 hours. The first five weeks already total \(8+11+7+12+10 = 48\) hours, so the sixth week must supply the remaining \(60-48 = 12\) hours. This matches choice D.

ANSWER 8: D

Problem 9:
The total payment is the three-digit multiple of 11 written \(1A2\). The multiples of 11 in that range are
\[
110,\ 121,\ 132,\ 143,\ 154.
\]
Each is divisible by 11 and the corresponding values of the digit \(A\) are 1, 2, 3, 4, 5. The only value among the given choices that actually occurs is therefore 2 (from 121). Hence A = 2.

ANSWER 9: B

Problem 10:
July has 31 days, each containing 24 hours, for a total of \(31 \times 24\) hours. The required average is therefore total rainfall divided by total hours:
\[
\frac{366}{31 \times 24}.
\]
This expression is exactly choice A.

ANSWER 10: A

Problem 11:
The daily banana counts form an arithmetic sequence of five terms whose first term is \(x\) and common difference is 6. Their sum is
\[
5x + (0+6+12+18+24) = 5x + 60 = 100 \implies 5x = 40 \implies x = 8.
\]
The fifth term is therefore \(8 + 4 \times 6 = 32\), matching choice D.

ANSWER 11: D

Problem 12:
Room A contributes a total age of \(6 \times 40 = 240\) person-years; room B contributes \(4 \times 25 = 100\) person-years. The combined group of 10 people therefore has total age 340, and the average age is
\[
\frac{340}{10} = 34.
\]
This matches choice D.

ANSWER 12: D

Problem 1:
What is being asked is to find the sum of the two smallest prime factors of 250. To do this, we need to factorize 250 into its prime factors.

First, let's break down 250 into its prime factors: 250 = 2 * 125. Further breaking down 125, we get 125 = 5 * 25. Then, breaking down 25 into 5 * 5, we have 250 = 2 * 5 * 5 * 5.

The prime factors of 250 are 2 and 5. The two smallest prime factors are 2 and 5. 

Now, we sum these two smallest prime factors: 2 + 5 = 7.

Let's examine the choices given:
A. 2 - This is one of the prime factors but not their sum.
B. 5 - This is also one of the prime factors but not their sum.
C. 7 - This is the sum of 2 and 5.
D. 10 - This is not the sum of the two smallest prime factors.
E. 12 - This is also not the sum of the two smallest prime factors.

The correct sum is 7.

ANSWER 1: C

Problem 2:
The task is to determine how many different four-digit numbers can be formed by rearranging the four digits in 2004.

The number 2004 has 2 zeros and 2 fours. To form different numbers, we need to consider the permutations of these digits.

The total number of permutations of 4 digits is 4!. However, since there are 2 identical zeros and 2 identical fours, we need to divide by 2! for each set of identical digits to correct for overcounting.

So, the total number of different arrangements = 4! / (2! * 2!) = 24 / (2 * 2) = 24 / 4 = 6.

Let's look at the choices:
A. 4 - This is less than the calculated permutations.
B. 6 - This matches our calculation.
C. 16 - This is more than the calculated permutations.
D. 24 - This is the total permutations without correcting for identical digits.
E. 81 - This is far more than the calculated permutations.

The correct number of different arrangements is 6.

ANSWER 2: B

Problem 3:
We need to determine which of the given expressions must be an odd integer, given that m and n are positive odd integers.

Let's examine each choice:
A. m + 3n: Since m is odd and 3n is odd (because n is odd and 3 is odd), their sum is even.
B. 3m - n: Here, 3m is odd (since m is odd) and n is odd, so their difference is even.
C. 3m^2 + 3n^2: Both 3m^2 and 3n^2 are odd (since m and n are odd, their squares are odd, and multiplying by 3 keeps them odd). The sum of two odd numbers is even.
D. (nm + 3)^2: nm is odd (product of two odd numbers), so nm + 3 is even (since 3 is odd), and the square of an even number is even.
E. 3mn: mn is odd (product of two odd numbers), and multiplying by 3 keeps it odd.

Let's verify the options with an example where m = 1 and n = 1:
A. 1 + 3*1 = 4 (even)
B. 3*1 - 1 = 2 (even)
C. 3*1^2 + 3*1^2 = 6 (even)
D. (1*1 + 3)^2 = 16 (even)
E. 3*1*1 = 3 (odd)

The expression that must be an odd integer is 3mn.

ANSWER 3: E

Problem 4:
The problem states that in a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

Let's denote the total number of marbles as T. The number of red marbles is T - 6, the number of green marbles is T - 8, and the number of blue marbles is T - 4.

The sum of the marbles of each color equals the total number of marbles: (T - 6) + (T - 8) + (T - 4) = T.

Simplifying this, we get 3T - 18 = T.

Rearranging the equation gives us 2T = 18.

Dividing both sides by 2, we find T = 9.

Let's check the choices:
A. 6
B. 8
C. 9
D. 10
E. 12

Our calculation matches choice C.

ANSWER 4: C

Problem 5:
We are asked to identify the largest number among the given choices.

Let's compare the numbers:
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089

First, we can eliminate D because it's less than 0.97. Now, comparing the rest:
- 0.97 is less than 0.979.
- 0.979 is greater than 0.9709.
- 0.9089 is less than 0.97 and thus less than 0.979.

So, the order from largest to smallest is: 0.979, 0.9709, 0.97, 0.9089, 0.907.

The largest number is 0.979.

ANSWER 5: B

Problem 6:
The aquarium has a base area of 100 cm * 40 cm = 4000 cm^2 and is filled with water to a depth of 37 cm. A rock with a volume of 1000 cm^3 is submerged.

The rise in water level is determined by the volume of the rock divided by the base area of the aquarium.

Rise = Volume of rock / Base area = 1000 cm^3 / 4000 cm^2 = 0.25 cm.

Let's look at the choices:
A. 0.25
B. 0.5
C. 1
D. 1.25
E. 2.5

Our calculation matches choice A.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We need to find the maximum number of principals during an 8-year period.

Let's consider the terms:
- The first principal serves from year 1 to 3.
- The second principal serves from year 4 to 6.
- The third principal serves from year 7 to 9.

However, the 8-year period ends before the third principal's term is over. So, we can have a principal starting their term at year 6 (overlapping with the second principal's last year) and serving until year 8 (and beyond).

Thus, the principals can serve as follows: 
1. Year 1-3
2. Year 4-6
3. Year 6-8 (starts before the end of the 8-year period)

This gives us 3 principals within the 8-year span, with some overlap.

Let's check the choices:
A. 2
B. 3
C. 4
D. 5
E. 8

Our analysis supports choice B, as having 4 principals would require a different arrangement that doesn't fit within the 8-year frame as neatly.

To maximize, consider: 
1st principal (1-3), 2nd (4-6), and then a new principal can start at the beginning of year 6 or just after, but for counting within the 8 years, we effectively count the principal starting in year 6 as the third.

However, upon closer inspection, we can actually fit in one more by having the sequence: 
1 (years 1-3), 2 (4-6), and then 3 can start at year 6 and serve till year 8, and a new principal (4th) can be considered to start right after, though not fully within the 8 years.

Actually, the maximum is achieved by having principals serve from years 1-3, 4-6, and then one more starting in year 6 or slightly after but counted within the period gives us 3. But considering year 6 as a starting poin

Problem 1:
We need the sum of the two smallest prime factors of 250.  
First, factor 250:  
250 = 2 × 125 = 2 × 5³.  
The prime factors are 2 and 5. The two smallest (and only) prime factors are 2 and 5.  
Sum = 2 + 5 = 7.  
ANSWER 1: C

Problem 2:
The digits of 2004 are 2, 0, 0, 4. We need four-digit numbers, so the first digit cannot be 0.  
Total arrangements of these four digits (treating the zeros as identical) = 4! / 2! = 12.  
Arrangements that start with 0: fix 0 as the first digit, then arrange the remaining 2, 0, 4 in 3! = 6 ways.  
Valid four-digit numbers = 12 − 6 = 6.  
ANSWER 2: B

Problem 3:
m and n are positive odd integers. Check each option for oddness:  
A. m + 3n: odd + (odd × odd) = odd + odd = even.  
B. 3m − n: odd − odd = even.  
C. 3m² + 3n²: odd² = odd; 3×odd = odd; odd + odd = even.  
D. (nm + 3)²: nm = odd; odd + 3 (odd) = even; even² = even.  
E. 3mn: odd × odd = odd; 3 × odd = odd.  
Only E must be odd.  
ANSWER 3: E

Problem 4:
Let r, g, b be the number of red, green, blue marbles.  
"All but 6 are red" → g + b = 6.  
"All but 8 are green" → r + b = 8.  
"All but 4 are blue" → r + g = 4.  
Add the three equations: (g+b) + (r+b) + (r+g) = 6 + 8 + 4 → 2(r+g+b) = 18 → r+g+b = 9.  
Total marbles = 9.  
ANSWER 4: C

Problem 5:
Compare the decimals by aligning digits:  
0.97  = 0.9700…  
0.979 = 0.9790…  
0.9709 = 0.9709…  
0.907 = 0.9070…  
0.9089 = 0.9089…  
The largest is 0.979.  
ANSWER 5: B

Problem 6:
Base area = 100 cm × 40 cm = 4000 cm².  
Volume of rock = 1000 cm³.  
Water level rise = volume / base area = 1000 / 4000 = 0.25 cm.  
ANSWER 6: A

Problem 7:
Each principal serves exactly 3 years. To maximize the number of different principals during an 8‑year period, we can have terms overlapping the boundaries.  
For example: a principal finishes during the first year, then two full 3‑year terms (6 years), and a new principal starts in the last year. That gives 4 principals.  
(If terms are placed end‑to‑end, 8 years can contain at most ⌈8/3⌉ = 3 full terms, but by shifting the 8‑year window we can intersect 4 terms.)  
ANSWER 7: C

Problem 8:
Target average = 10 hours/week for 6 weeks → total hours needed = 10 × 6 = 60.  
Hours worked in first 5 weeks = 8 + 11 + 7 + 12 + 10 = 48.  
Hours needed in final week = 60 − 48 = 12.  
ANSWER 8: D

Problem 9:
Total paid = $1A2 = 100 + 10A + 2 = 102 + 10A.  
This is divisible by 11 (11 members each paid an integer amount).  
102 mod 11 = 3 (since 11×9=99). 10A mod 11 = −A (since 10 ≡ −1).  
So 3 − A ≡ 0 (mod 11) → A ≡ 3 (mod 11).  
A is a digit 0–9, so A = 3. (Check: 132 ÷ 11 = 12.)  
ANSWER 9: D

Problem 10:
July has 31 days. Total hours in July = 31 × 24.  
Total rainfall = 366 inches.  
Average rainfall per hour = 366 / (31 × 24).  
ANSWER 10: A

Problem 11:
Let bananas eaten on May 1 = x.  
May 2 = x+6, May 3 = x+12, May 4 = x+18, May 5 = x+24.  
Sum = 5x + (0+6+12+18+24) = 5x + 60 = 100.  
5x = 40 → x = 8.  
May 5 = 8 + 24 = 32.  
ANSWER 11: D

Problem 12:
Room A: 6 people, average 40 → total age = 6 × 40 = 240.  
Room B: 4 people, average 25 → total age = 4 × 25 = 100.  
Combined total age = 240 + 100 = 340.  
Combined number of people = 6 + 4 = 10.  
Combined average = 340 ÷ 10 = 34.  
ANSWER 12: D

Problem 1:
We need the two smallest prime factors of 250.
First, find the prime factorization of 250:
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5 = 2 × 5³
The prime factors are 2 and 5.
The two smallest (and only) prime factors are 2 and 5.
Their sum is 2 + 5 = 7.
ANSWER 1: C

Problem 2:
The digits in 2004 are: 2, 0, 0, 4.
Total arrangements of these 4 digits (with 0 repeated twice) = 4! / 2! = 24 / 2 = 12.
However, a four-digit number cannot start with 0. Let's count arrangements starting with 0:
Fix 0 in the first position. The remaining digits are 2, 0, 4 (all distinct).
Arrangements = 3! = 6.
Valid four-digit numbers = 12 − 6 = 6.
Listing them to verify: 2004, 2040, 2400, 4002, 4020, 4200. That's exactly 6.
ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we test each option:
A. m + 3n = odd + 3(odd) = odd + odd = even ✗
B. 3m − n = 3(odd) − odd = odd − odd = even ✗
C. 3m² + 3n² = 3(odd²) + 3(odd²) = 3(odd) + 3(odd) = odd + odd = even ✗
D. (nm + 3)² = (odd·odd + 3)² = (odd + odd)² = (even)² = even ✗
E. 3mn = 3 × odd × odd = odd × odd × odd = odd ✓
The product of any number of odd integers is always odd.
ANSWER 3: E

Problem 4:
Let R, G, B be the counts of red, green, and blue marbles. Let T = R + G + B.
"All but 6 are red" means the non-red marbles total 6: G + B = 6
"All but 8 are green" means the non-green marbles total 8: R + B = 8
"All but 4 are blue" means the non-blue marbles total 4: R + G = 4
Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
R + G + B = 9
ANSWER 4: C

Problem 5:
Line up all numbers with four decimal places for easy comparison:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
Comparing digit by digit from left to right:
- The tenths digit: A, B, C all have 9; D and E have 9. All tied.
- The hundredths digit: A, B, C have 7; D, E have 0. So D and E are eliminated.
- The thousandths digit: A has 0, B has 9, C has 0. B wins with 9.
So 0.979 is the largest.
ANSWER 5: B

Problem 6:
The base area of the aquarium = 100 cm × 40 cm = 4000 cm².
When the rock (volume = 1000 cm³) is submerged, it displaces 1000 cm³ of water.
This displaced water spreads over the base area, causing the level to rise.
Rise in water level = Volume of rock / Base area = 1000 / 4000 = 0.25 cm.
Check: new depth = 37 + 0.25 = 37.25 cm < 50 cm, so no overflow.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. We want to maximize the number of principals during an 8-year period.
To maximize, we want terms to overlap the 8-year window as efficiently as possible:
- Principal 1: term ends on the very first day of the 8-year period (counts as serving during the period).
- Principal 2: serves a full 3-year term starting on day 1.
- Principal 3: serves the next full 3-year term.
- Principal 4: starts a 3-year term at the beginning of year 7; part of this term falls within the 8-year period.
Timeline: P1(end) | P1(3 yrs) | P2(3 yrs) | P3(partial)
That's 0 + 3 + 3 = 6 years of full terms, plus partial terms at both ends.
Could we fit a 5th? After Principal 4 starts at year 7, their term ends at year 10, which is past the 8-year window. A 5th principal would start at year 10, outside the period. Before Principal 1, the previous principal's term would end before the period starts.
Maximum = 4 principals.
ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours/week over 6 weeks.
Total hours needed = 10 × 6 = 60 hours.
Hours worked in the first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
ANSWER 8: D

Problem 9:
The total amount paid is $1A2, and 11 members each paid the same integer amount.
So the 3-digit number 1A2 must be divisible by 11.
Using the divisibility rule for 11 (alternating sum of digits):
1 − A + 2 = 3 − A
For divisibility by 11, we need 3 − A = 0, 11, −11, ...
3 − A = 0 → A = 3
3 − A = 11 → A = −8 (not a digit)
3 − A = −11 → A = 14 (not a digit)
So A = 3. Check: 132 ÷ 11 = 12. Each member paid $12. ✓
ANSWER 9: D

Problem 10:
We need the average rainfall in inches per hour during July 1861.
Total rainfall = 366 inches.
July has 31 days, and each day has 24 hours.
Total hours in July = 31 × 24.
Average rainfall per hour = Total rainfall / Total hours = 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days (May 1–5), eating 6 more each day than the previous day.
Let x = bananas eaten on May 1.
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
Sum = x + (x+6) + (x+12) + (x+18) + (x+24) = 5x + 60
Set equal to 100: 5x + 60 = 100 → 5x = 40 → x = 8
Bananas on May 5 = x + 24 = 8 + 24 = 32.
ANSWER 11: D

Problem 12:
Room A: 6 people with average age 40 → total age = 6 × 40 = 240
Room B: 4 people with average age 25 → total age = 4 × 25 = 100
Combined: 6 + 4 = 10 people, total age = 240 + 100 = 340
Average age = 340 / 10 = 34.
ANSWER 12: D

Problem 1:
We need the sum of the two smallest prime factors of 250.
First factor 250:
250 = 2 × 125
125 = 5 × 25 = 5 × 5 × 5
So 250 = 2 × 5³.
The distinct prime factors are 2 and 5.
The two smallest are 2 and 5.
Their sum is 2 + 5 = 7.
(Notice that 7 is not a factor, but it is the sum; choices A and B are the individual factors, while D is their product.)

ANSWER 1: C

Problem 2:
We must form different four-digit numbers by rearranging the digits 2, 0, 0, 4.
A four-digit number cannot start with 0.
Total distinct permutations of the four digits (with the two 0s identical) are 4!/2! = 12.
Now subtract the permutations that begin with 0.
If 0 is fixed in the thousands place, we arrange 2, 0, 4 in the remaining three spots: 3! = 6 such arrangements (e.g., 0204, 0240, 0024, 0042, 0402, 0420), none of which are valid four-digit numbers.
Valid numbers = 12 − 6 = 6.
Listing them confirms: 2004, 2040, 2400, 4002, 4020, 4200.
(24 would be 4! if all digits were distinct and nonzero; 16 and 4 are under/over counts.)

ANSWER 2: B

Problem 3:
Given m and n are positive odd integers, we check the parity of each choice.
A. m + 3n: odd + 3(odd) = odd + odd = even.
B. 3m − n: odd − odd = even.
C. 3m² + 3n²: m² and n² are odd, so 3(odd) + 3(odd) = odd + odd = even.
D. (nm + 3)²: nm is odd·odd = odd; odd + 3 = even; even² = even.
E. 3mn: 3 is odd, m is odd, n is odd. The product of odd numbers is odd.
Only choice E is guaranteed to be odd.

ANSWER 3: E

Problem 4:
Let R, G, B be the numbers of red, green, and blue marbles.
“All but 6 are red” means non-red marbles total 6: G + B = 6.
“All but 8 are green” means non-green marbles total 8: R + B = 8.
“All but 4 are blue” means non-blue marbles total 4: R + G = 4.
Add the three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2(R + G + B) = 18
Total marbles = R + G + B = 9.
(Checking: R = 3, G = 1, B = 5, and 1+5=6, 3+5=8, 3+1=4.)

ANSWER 4: C

Problem 5:
Compare the decimals by aligning place values:
A. 0.9700
B. 0.9790
C. 0.9709
D. 0.9070
E. 0.9089
All start with 0.9. In the hundredths place, A, B, C have 7 while D, E have 0, so D and E are out.
Among A, B, C, look at the thousandths place: A has 0, B has 9, C has 0. Thus 0.979 is the largest.

ANSWER 5: B

Problem 6:
The rock is completely submerged, so it displaces a volume of water equal to its own volume, 1000 cm³.
The base of the aquarium is 100 cm × 40 cm, giving a base area of 4000 cm².
The water level rises by the displaced volume divided by the base area:
Rise = 1000 / 4000 = 0.25 cm.
Check for overflow: new depth = 37 + 0.25 = 37.25 cm, well below the 50 cm height, so the rise stays 0.25 cm.

ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To maximize the number of principals whose terms overlap an 8-year period, align the period so it catches the very end of one term and the very beginning of another.
Example timeline (years are inclusive blocks):
Principal 1: years 1–3 (period catches year 3)
Principal 2: years 4–6 (fully inside)
Principal 3: years 7–9 (fully inside)
Principal 4: years 10–12 (period catches year 10)
If the 8-year period is year 3 through year 10, it overlaps with all four principals.
Can we get 5? Five sequential 3-year terms span at least 12 − 1 = 11 years from the end of the first to the start of the fifth, which exceeds 8, so 5 is impossible.
Thus the maximum is 4.

ANSWER 7: C

Problem 8:
Theresa needs an average of 10 hours per week over 6 weeks.
Total hours required = 10 × 6 = 60 hours.
Hours already worked: 8 + 11 + 7 + 12 + 10 = 48 hours.
Hours needed in the final week = 60 − 48 = 12 hours.
(If she worked 10, her average would be 58/6 < 10; 13 would give 61/6 > 10 but more than needed.)

ANSWER 8: D

Problem 9:
Eleven members each paid the same integer amount, so the total 1A2 must be divisible by 11.
Using the divisibility rule for 11: (sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11.
For 1A2: (1 + 2) − A = 3 − A.
We need 3 − A ≡ 0 (mod 11), so A ≡ 3 (mod 11).
Since A is a single digit, A = 3.
Check: 132 ÷ 11 = 12, an integer.
(Other digits fail: 102/11 ≈ 9.27, 112/11 ≈ 10.18, etc.)

ANSWER 9: D

Problem 10:
Average rainfall in inches per hour = total inches / total hours.
Total inches = 366.
July always has 31 days.
Total hours = 31 days × 24 hours/day.
So the average is 366 / (31 × 24).
(The other choices either invert the ratio or multiply instead of divide.)

ANSWER 10: A

Problem 11:
Let x be the number of bananas eaten on May 1.
Then the daily amounts are:
May 1: x
May 2: x + 6
May 3: x + 12
May 4: x + 18
May 5: x + 24
Total over 5 days: 5x + (0 + 6 + 12 + 18 + 24) = 5x + 60 = 100.
Solving: 5x = 40 → x = 8.
On May 5: x + 24 = 8 + 24 = 32.
(20 would be the average 100/5, but the sequence is increasing, so the last day must be larger than the average.)

ANSWER 11: D

Problem 12:
Total age in Room A = 6 people × 40 years/person = 240 years.
Total age in Room B = 4 people × 25 years/person = 100 years.
Combined total age = 240 + 100 = 340 years.
Combined total people = 6 + 4 = 10.
Average age = 340 / 10 = 34 years.
(The simple average of 40 and 25 is 32.5, but the groups have different sizes, so the correct weighted average is 34.)

ANSWER 12: D

Problem 1:
We need to find the sum of the two smallest prime factors of 250. First, let's find the prime factorization of 250.
250 = 10 × 25
250 = (2 × 5) × (5 × 5) = 2 × 5³
The prime factors of 250 are 2 and 5. Since these are the only prime factors, they are also the two smallest.
Their sum is 2 + 5 = 7.
Looking at the choices, 7 is option C. (Option D, 10, might be tempting if someone adds 5+5, but 5 is only counted once as a prime factor).
ANSWER 1: C

Problem 2:
We want to find how many different four-digit numbers can be formed by rearranging the digits of 2004. The digits are 2, 0, 0, and 4.
If we were arranging 4 distinct digits, there would be 4! = 24 permutations. However, the two 0s are identical, so we must divide by the number of ways to arrange them (2!):
Total permutations = 4! / 2! = 24 / 2 = 12.
But a four-digit number cannot start with 0. Let's subtract the arrangements that start with 0.
If the first digit is 0, we are arranging the remaining digits (2, 0, 4), which can be done in 3! / 1! = 6 ways.
So, the number of valid four-digit numbers is 12 - 6 = 6.
(These are 2004, 2040, 2400, 4002, 4020, 4200).
ANSWER 2: B

Problem 3:
We are given that m and n are positive odd integers, and we need to determine which expression must also be an odd integer. Let's evaluate each option using the properties of odd and even numbers (odd + odd = even, odd × odd = odd, etc.):
A. m + 3n: m is odd, 3n is odd (odd × odd = odd). odd + odd = even. (Not odd)
B. 3m − n: 3m is odd, n is odd. odd − odd = even. (Not odd)
C. 3m² + 3n²: m² is odd, 3m² is odd; n² is odd, 3n² is odd. odd + odd = even. (Not odd)
D. (nm + 3)²: nm is odd, nm + 3 is even (odd + odd = even). even² = even. (Not odd)
E. 3mn: 3, m, and n are all odd. odd × odd × odd = odd. (Must be odd!)
ANSWER 3: E

Problem 4:
Let R, G, and B be the number of red, green, and blue marbles, respectively. Let T be the total number of marbles, so T = R + G + B.
We are given:
- "All but 6 are red" means T - R = 6, so G + B = 6.
- "All but 8 are green" means T - G = 8, so R + B = 8.
- "All but 4 are blue" means T - B = 4, so R + G = 4.
Adding these three equations together:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2(R + G + B) = 18
2T = 18
T = 9.
There are 9 marbles in the jar. (Checking: R=3, G=1, B=5, which sums to 9 and satisfies all conditions).
ANSWER 4: C

Problem 5:
To find the largest number, we compare the digits from left to right (tenths, hundredths, thousandths):
A. 0.97
B. 0.979
C. 0.9709
D. 0.907
E. 0.9089
First, look at the tenths place: A, B, and C have a 9, while D and E have a 0. So D and E are out.
Now compare A, B, and C to the hundredths place: all have a 7.
Now compare to the thousandths place: A has an implied 0 (0.970), B has a 9, and C has a 0.
Since 9 > 0, B is the largest number.
ANSWER 5: B

Problem 6:
When the rock is submerged, it displaces a volume of water equal to its own volume (1000 cm³). This displaced water spreads out over the base of the aquarium, causing the water level to rise.
The base of the aquarium measures 100 cm by 40 cm, so the area of the base is:
Area = 100 cm × 40 cm = 4000 cm².
The rise in the water level (height) is the volume of the displaced water divided by the base area:
Rise = Volume / Area = 1000 cm³ / 4000 cm² = 1/4 cm = 0.25 cm.
ANSWER 6: A

Problem 7:
Each principal serves exactly one 3-year term. To find the maximum number of principals in an 8-year period, we want to fit as many 3-year terms as possible into the 8 years.
If a principal starts at the very beginning of the 8-year period, their term covers years 1, 2, and 3.
The next principal could start at the beginning of year 4, covering years 4, 5, and 6.
A third principal could start at the beginning of year 7, covering years 7, 8, and 9. Since their term overlaps with the 8-year period (during years 7 and 8), they count as a principal during this time.
Thus, the maximum number of principals the school could have during an 8-year period is 3. (4 principals would require at least 4 × 3 = 12 years).
ANSWER 7: B

Problem 8:
Theresa needs an average of 10 hours per week for 6 weeks. The total number of hours she must work is:
Total hours = 10 hours/week × 6 weeks = 60 hours.
For the first 5 weeks, she worked:
8 + 11 + 7 + 12 + 10 = 48 hours.
To find how many hours she must work in the final week, subtract the hours she has already worked from the total required:
Hours needed = 60 - 48 = 12 hours.
ANSWER 8: D

Problem 9:
The 11 members each paid the same integer amount, meaning the total amount paid, $1A2, must be divisible by 11.
To test for divisibility by 11, we find the alternating sum of the digits: subtract the second digit from the sum of the first and third digits.
(1 + 2) - A = 3 - A.
For the number to be divisible by 11, the alternating sum must be a multiple of 11 (including 0). Since A is a single digit from 0 to 9, the only way 3 - A can be a multiple of 11 is if it equals 0.
3 - A = 0
A = 3.
Let's verify: if A = 3, the total is $132. Dividing $132 by 11 members gives $12 per member, which is an integer amount.
ANSWER 9: D

Problem 10:
We are looking for the average rainfall in inches per hour.
Average = (Total rainfall) / (Total number of hours).
The total rainfall is 366 inches.
The month of July has 31 days. Since there are 24 hours in a day, the total number of hours in July is 31 × 24.
Therefore, the average rainfall per hour is 366 / (31 × 24).
ANSWER 10: A

Problem 11:
Big Al ate 100 bananas over 5 days, eating 6 more bananas each day than the previous day. Let x be the number of bananas he ate on the first day.
The number of bananas eaten on the 5 days are:
Day 1: x
Day 2: x + 6
Day 3: x + 12
Day 4: x + 18
Day 5: x + 24
The sum of these is 100:
x + (x + 6) + (x + 12) + (x + 18) + (x + 24) = 100
5x + 60 = 100
5x = 40
x = 8.
He ate 8 bananas on the first day. We want to know how many he ate on May 5 (the 5th day):
Day 5 = x + 24 = 8 + 24 = 32 bananas.
ANSWER 11: D

Problem 12:
To find the combined 

**Problem 1:**  
Factor 250: \(250 = 2 \times 5^3\).  
The prime factors are \(2\) and \(5\).  
Sum of the two smallest prime factors: \(2 + 5 = 7\).  

**Answer:** C  

---

**Problem 2:**  
Digits in 2004 are \(\{2,0,0,4\}\).  
Total permutations of the four digits (with two zeros identical) is \(\dfrac{4!}{2!}=12\).  
Numbers beginning with 0 are not four‑digit numbers; fix one zero in the first place and permute the remaining three digits \(\{2,0,4\}\): \(3! = 6\) such numbers.  
Valid four‑digit numbers = \(12 - 6 = 6\).  

**Answer:** B  

---

**Problem 3:**  
Let \(m = 2a+1\) and \(n = 2b+1\) (both odd).  

- \(m+3n = 2a+1 + 6b+3 = 2(a+3b+2)\) → even  
- \(3m-n = 6a+3 -2b-1 = 2(3a-b+1)\) → even  
- \(3m^2+3n^2 = 3[(2a+1)^2+(2b+1)^2]\) → \(m^2,n^2\) odd, their sum even ⇒ factor 3 gives even  
- \((nm+3)^2 = (\text{odd}+3)^2 = \text{even}^2 = \text{even}\)  
- \(3mn = 3(\text{odd}\times\text{odd}) = 3(\text{odd})\) → odd  

Only choice **E** must be odd.  

**Answer:** E  

---

**Problem 4:**  
Let \(T\) be the total number of marbles.  
- Red marbles: \(T-6\)  
- Green marbles: \(T-8\)  
- Blue marbles: \(T-4\)  

Thus  
\[
T = (T-6)+(T-8)+(T-4) = 3T - 18 \quad\Longrightarrow\quad 2T = 18 \;\Rightarrow\; T = 9.
\]

**Answer:** C  

---

**Problem 5:**  
Compare the decimals:  

\[
0.979 > 0.9709 > 0.97 > 0.9089 > 0.907.
\]

The largest number is \(0.979\).  

**Answer:** B  

---

**Problem 6:**  
Base area \(=100\text{ cm}\times40\text{ cm}=4000\text{ cm}^2\).  

Increase in water volume caused by the rock = rock volume = \(1000\text{ cm}^3\).  

Rise in water level \(h\) satisfies \(4000h = 1000 \;\Rightarrow\; h = \frac{1000}{4000}=0.25\) cm.  

**Answer:** A  

---

**Problem 7:**  
A principal serves exactly a 3‑year term.  
In an 8‑year span the following schedule is possible:

- Principal 1: years 0‑3  
- Principal 2: years 3‑6  
- Principal 3: years 6‑9  

All three serve at least a portion of the 8‑year interval. No fourth principal can start before year 9, which lies outside the interval. Hence the maximum number of principals is **3**.  

**Answer:** B  

---

**Problem 8:**  
Total required hours for 6 weeks at 10 h/week: \(6 \times 10 = 60\) hours.  
Hours already logged: \(8+11+7+12+10 = 48\) hours.  

Hours needed in the final week: \(60 - 48 = 12\) hours.  

**Answer:** D  

---

**Problem 9:**  
The total paid is the three‑digit number \(1A2 = 100 + 10A + 2 = 102 + 10A\).  

Since 11 members paid the same integer amount, the total must be divisible by 11:  

\[
102 + 10A \equiv 0 \pmod{11}.
\]

Because \(10 \equiv -1 \pmod{11}\),

\[
102 + 10A \equiv 3 - A \equiv 0 \pmod{11} \;\Longrightarrow\; A \equiv 3 \pmod{11}.
\]

With \(A\) a single digit, \(A = 3\).  

**Answer:** D  

---

**Problem 10:**  
July has 31 days, each day 24 h, so total hours = \(31 \times 24\).  

Average rainfall = \(\dfrac{366\text{ in}}{31 \times 24\text{ h}} = \frac{366}{31 \times 24}\).  

**Answer:** A  

---

**Problem 11:**  
Let the number of bananas eaten on May 1 be \(a\).  
Then the daily amounts are \(a,\; a+6,\; a+12,\; a+18,\; a+24\).  

Sum for 5 days:  
\[
5a + (6+12+18+24) = 5a + 60 = 100 \;\Longrightarrow\; 5a = 40 \;\Longrightarrow\; a = 8.
\]

Bananas on May 5: \(a+24 = 8+24 = 32\).  

**Answer:** D  

---

**Problem 12:**  
Room A total age = \(6 \times 40 = 240\).  
Room B total age = \(4 \times 25 = 100\).  

Combined total age = \(240 + 100 = 340\).  
Combined number of people = \(6 + 4 = 10\).  

Average age = \(\dfrac{340}{10} = 34\).  

**Answer:** D  

---

**Final answers**

ANSWER 1: C  
ANSWER 2: B  
ANSWER 3: E  
ANSWER 4: C  
ANSWER 5: B  
ANSWER 6: A  
ANSWER 7: B  
ANSWER 8: D  
ANSWER 9: D  
ANSWER 10: A  
ANSWER 11: D  
ANSWER 12: D

Problem 1:  
We are asked for the sum of the two smallest prime factors of 250.  
First, factorize 250:  
250 = 2 × 125 = 2 × 5 × 25 = 2 × 5 × 5 × 5.  
The prime factors are 2 and 5.  
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.  
Tempting wrong choices: A (2 is the smallest prime factor, but not the sum), B (5 is the next, but not the sum).  
ANSWER 1: C  

Problem 2:  
We are to find how many different four-digit numbers can be formed by rearranging the digits in 2004.  
The digits are 2, 0, 0, 4. Since there are two identical zeros, the total permutations are 4! / 2! = 24 / 2 = 12, but not all are valid four-digit numbers because numbers cannot start with 0.  
Valid permutations: The first digit must be 2 or 4.  
- If first digit is 2: Remaining digits 0, 0, 4. The permutations are 2004, 2040, 2400 → 3 numbers.  
- If first digit is 4: Remaining digits 0, 0, 2. The permutations are 4002, 4020, 4200 → 3 numbers.  
Total valid numbers: 3 + 3 = 6.  
Tempting wrong choices: A (4 ignores duplicate zeros), C (16 ignores leading zero rule), D (24 ignores duplicates and leading zero), E (81 is 3^4, not applicable).  
ANSWER 2: B  

Problem 3:  
We are given that m and n are positive odd integers. We need to find which expression must also be odd.  
Recall: odd × odd = odd, odd + odd = even, even + even = even, odd + even = odd.  
Check each option:  
A. m + 3n: m is odd, 3n is odd (since 3 is odd and n is odd), so odd + odd = even.  
B. 3m − n: 3m is odd, n is odd, so odd − odd = even.  
C. 3m² + 3n²: m² is odd, 3m² is odd, similarly 3n² is odd, so odd + odd = even.  
D. (nm + 3)²: nm is odd (odd × odd), nm + 3 is even (odd + odd), and (even)² is even.  
E. 3mn: mn is odd, 3mn is odd (odd × odd).  
Thus, only E is odd.  
Tempting wrong choices: A, B, C, D are even as shown.  
ANSWER 3: E  

Problem 4:  
We are given:  
- All but 6 are red → green + blue = 6.  
- All but 8 are green → red + blue = 8.  
- All but 4 are blue → red + green = 4.  
Let R, G, B be the number of red, green, and blue marbles.  
We have:  
G + B = 6  
R + B = 8  
R + G = 4  
Add all three equations: 2R + 2G + 2B = 18 → R + G + B = 9.  
Thus, total marbles = 9.  
Tempting wrong choices: A (6 is G + B), B (8 is R + B), C (9 is correct), D (10), E (12).  
ANSWER 4: C  

Problem 5:  
We are to find the largest number among:  
A. 0.97  
B. 0.979  
C. 0.9709  
D. 0.907  
E. 0.9089  
Compare digit by digit:  
- All start with 0.9.  
- Next digit: A, B, C have 7, D and E have 0 (so D and E are smaller).  
- Among A, B, C: A is 0.9700, B is 0.9790, C is 0.9709.  
- B is larger than A and C because 9 > 7 in the third decimal place.  
Thus, B is the largest.  
Tempting wrong choices: A (0.97 is less than 0.979), C (0.9709 < 0.979), D and E are smaller.  
ANSWER 5: B  

Problem 6:  
The aquarium has a base of 100 cm × 40 cm and height 50 cm. It is filled to a depth of 37 cm. A rock of volume 1000 cm³ is submerged. We need to find the rise in water level.  
The base area is 100 × 40 = 4000 cm².  
The volume of water displaced by the rock is 1000 cm³.  
The rise in water level is volume / base area = 1000 / 4000 = 0.25 cm.  
Tempting wrong choices: A (0.25 is correct), B (0.5 is twice as much), C (1 is too high), D (1.25), E (2.5).  
ANSWER 6: A  

Problem 7:  
Each principal serves a 3-year term. We need the maximum number of principals in an 8-year period.  
To maximize, we can have principals serving in overlapping terms:  
- Principal 1: Years 1-3  
- Principal 2: Years 2-4  
- Principal 3: Years 5-7  
- Principal 4: Years 6-8  
This gives 4 principals.  
Alternatively, we could have:  
- Principal 1: Years 1-3  
- Principal 2: Years 4-6  
- Principal 3: Years 7-9 (but 9 is outside 8 years, so not possible)  
But with overlapping, we get 4.  
Tempting wrong choices: A (2 is too few), B (3 is possible but not maximum), C (4 is maximum), D (5), E (8 is impossible).  
ANSWER 7: C  

Problem 8:  
Theresa needs an average of 10 hours per week for 6 weeks. Total hours needed: 10 × 6 = 60 hours.  
She has already worked: 8, 11, 7, 12, 10 hours in the first 5 weeks.  
Sum of first 5 weeks: 8 + 11 + 7 + 12 + 10 = 48 hours.  
Hours needed in the final week: 60 - 48 = 12 hours.  
Tempting wrong choices: A (9 is too few), B (10 is average but not what is needed), C (11 is close but not enough), D (12 is correct), E (13 is too many).  
ANSWER 8: D  

Problem 9:  
The total amount paid is $1A2, which is a three-digit number. Since 11 members paid the same integer amount, the total must be divisible by 11.  
The number is 1A2, which in numerical form is 100 + 10A + 2 = 102 + 10A.  
This must be divisible by 11: 102 + 10A ≡ 0 mod 11.  
Compute 102 mod 11: 11 × 9 = 99, so 102 ≡ 3 mod 11.  
Thus, 3 + 10A ≡ 0 mod 11 → 10A ≡ -3 ≡ 8 mod 11.  
Since 10 ≡ -1 mod 11, we have -A ≡ 8 mod 11 → A ≡ -8 ≡ 3 mod 11.  
Thus, A = 3.  
Tempting wrong choices: A (0), B (1), C (2), D (3 is correct), E (4).  
ANSWER 9: D  

Problem 10:  
In July 1861, 366 inches of rain fell. July has 31 days, so total hours = 31 × 24.  
Average rainfall per hour = total rain / total hours = 366 / (31 × 24).  
Tempting wrong choices: A (correct), B (multiplies by 31), C (multiplies by 24), D (inverts), E (multiplies all).  
ANSWER 10: A  

Problem 11:  
Big Al ate 100 bananas from May 1 to May 5 (5 days). Each day he ate 6 more than the previous day.  
Let x be the number eaten on May 1.  
Then:  
Day 1: x  
Day 2: x + 6  
Day 3: x + 12  
Day 4: x + 18  
Day 5: x + 24  
Total: 5x + 60 = 100 → 5x = 40 → x = 8.  
On May 5: x + 24 = 8 + 24 = 32.  
Tempting wrong choices: A (20 is too low), B (22), C (30), D (32 is correct), E (34).  
ANSWER 11: D  

Problem 12:  
Average age of 6 people in Room A is 40. Total age in Room A: 6 × 40 = 240.  
Average age of 4 people in Room B is 25. Total age in Room B: 4 × 25 = 100.  
Combined total age: 240 + 100 = 340.  
Combined number of people: 6 + 4 = 10.  
Average age: 340 / 10 = 34.  
Tempting wrong choices: A (32.5),

### Problem 1:
What is the sum of the two smallest prime factors of 250?
First, we calculate the prime factorization of 250: 250 is even, so the first prime factor is 2. 250 ÷ 2 = 125 = 5³. The full prime factorization is 2 × 5³. The only prime factors of 250 are 2 and 5, the two smallest. Their sum is 2 + 5 = 7. Tempting wrong answers include B (5, which is only one prime factor) and D (10, the product of the primes).
ANSWER 1: C

---

### Problem 2:
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
The digits are 2, 0, 0, 4. A valid four-digit number cannot start with 0, so the thousands place only has 2 choices: 2 or 4. If the thousands digit is 2, the remaining digits are 0,0,4, which form 3 unique numbers: 2004, 2040, 2400. If the thousands digit is 4, the remaining digits are 0,0,2, which form 3 more unique numbers: 4002, 4020, 4200. Total unique four-digit numbers: 6. Tempting wrong answers include D (24, the permutation of 4 distinct digits, ignoring duplicate 0s and leading zero rule).
ANSWER 2: B

---

### Problem 3:
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Use odd/even arithmetic rules: odd×odd=odd, odd+odd=even, odd-odd=even. Test each option:
- A. m + 3n: odd + odd = even ❌
- B. 3m - n: odd - odd = even ❌
- C. 3m² + 3n²: odd + odd = even ❌
- D. (nm + 3)²: (odd + odd)² = even² = even ❌
- E. 3mn: odd×odd×odd=odd ✅
ANSWER 3: E

---

### Problem 4:
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Let total marbles = T, R=red, G=green, B=blue. "All but X are Y" means non-Y marbles = X, so:
G+B=6, R+B=8, R+G=4. Add all three equations: 2(R+G+B)=18 → 2T=18 → T=9. Verify: R=3, G=1, B=5, sum to 9, which matches all conditions. Tempting wrong answer E (18, the unadjusted sum of the non-marble counts).
ANSWER 4: C

---

### Problem 5:
Which of the following numbers is the largest?
Rewrite all decimals to 4 places to compare: A=0.9700, B=0.9790, C=0.9709, D=0.9070, E=0.9089. The largest value is 0.9790 (option B). Tempting wrong answer C (0.9709, which is smaller than 0.979).
ANSWER 5: B

---

### Problem 6:
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
The base area of the aquarium is 100×40=4000 cm². The volume of the rock equals the volume of displaced water, so base area × height rise = rock volume. Solve for height rise: 1000 ÷ 4000 = 0.25 cm. No overflow occurs (37 + 0.25 < 50), so the rise is 0.25 cm.
ANSWER 6: A

---

### Problem 7:
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
To maximize the number of principals, stagger their terms to fit as many as possible into the 8-year window. For example, in the period 2000-2007: Principal 1 serves until the end of 2000, Principal 2 serves 2001-2003, Principal 3 serves 2004-2006, Principal 4 starts in 2007. This gives 4 principals, the maximum possible. A 5th principal would require an extra 3 years beyond the 8-year window. Tempting wrong answer B (3, the minimum number of principals, not maximum).
ANSWER 7: C

---

### Problem 8:
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work for the final week to earn the tickets?
Total required hours for 6 weeks: 6×10=60. Sum of first 5 weeks: 8+11+7+12+10=48. Required hours for the final week: 60-48=12.
ANSWER 8: D

---

### Problem 9:
Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?
The total cost 102 + 10A must be divisible by 11. The divisibility rule for 11 states that the alternating sum of digits must be divisible by 11: 1 - A + 2 = 3 - A, which equals 0 (the only valid multiple of 11 for a digit A) when A=3. Verify: 132 ÷ 11 = 12, an integer.
ANSWER 9: D

---

### Problem 10:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
July has 31 days, with 24 hours per day, so total hours in July = 31×24. Average rainfall per hour = total inches ÷ total hours = 366/(31×24).
ANSWER 10: A

---

### Problem 11:
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Let x = bananas eaten on May 1. The amounts for each day are x, x+6, x+12, x+18, x+24. Sum: 5x + 60 = 100 → 5x=40 → x=8. Bananas eaten on May 5: 8+24=32. Verify sum: 8+14+20+26+32=100.
ANSWER 11: D

---

### Problem 12:
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Total age of Room A: 6×40=240. Total age of Room B:4×25=100. Combined total age: 340, total people:10. Combined average: 340÷10=34.
ANSWER 12: D

### Problem 1:
We need the sum of the two smallest prime factors of 250.  
First factorize 250:  
$250 = 25 \times 10 = 5^2 \times 2 \times 5 = 2 \times 5^3$.  
The prime factors are 2, 5, 5, 5. The two smallest are 2 and 5.  
Sum: $2 + 5 = 7$.  
ANSWER 1: C

---

### Problem 2:
We need the number of distinct 4-digit numbers from digits 2, 0, 0, 4.  
A 4-digit number cannot start with 0.  
- If the first digit is 2: remaining digits are 0, 0, 4. The number of distinct arrangements is $\frac{3!}{2!} = 3$ (2004, 2040, 2400).  
- If the first digit is 4: remaining digits are 0, 0, 2. Similarly, $\frac{3!}{2!} = 3$ arrangements (4002, 4020, 4200).  
Total: $3 + 3 = 6$.  
ANSWER 2: B

---

### Problem 3:
We need which expression is always odd when $m,n$ are positive odd integers.  
Recall: odd × odd = odd, odd + odd = even, even × any = even, odd² = odd.  
- **A. $m + 3n$**: $m$ (odd) + $3n$ (odd×odd=odd) = even.  
- **B. $3m - n$**: $3m$ (odd) - $n$ (odd) = even.  
- **C. $3m^2 + 3n^2$**: $3m^2$ (odd×odd=odd) + $3n^2$ (odd) = even.  
- **D. $(nm + 3)^2$**: $nm$ (odd×odd=odd) + 3 (odd) = even; even² = even.  
- **E. $3mn$**: 3 (odd) × $m$ (odd) × $n$ (odd) = odd.  
ANSWER 3: E

---

### Problem 4:
Let $R$=red, $G$=green, $B$=blue, total $T=R+G+B$.  
- "All but 6 are red": $G+B=6$ → $T = R +6$ → $R=T-6$.  
- "All but 8 are green": $R+B=8$ → $T=G+8$ → $G=T-8$.  
- "All but 4 are blue": $R+G=4$ → $T=B+4$ → $B=T-4$.  
Substitute into $T=R+G+B$:  
$T = (T-6)+(T-8)+(T-4) = 3T - 18$  
$2T=18$ → $T=9$.  
ANSWER 4: C

---

### Problem 5:
Compare the decimals digit by digit:  
- All have 0.9 as the first two digits.  
- Third decimal place: A=7, B=7, C=7, D=0, E=0 → D/E are smaller.  
- Fourth decimal place: A=0 (0.9700), B=9 (0.9790), C=0 (0.9709). B is largest.  
ANSWER 5: B

---

### Problem 6:
The volume of the submerged rock equals the volume of displaced water.  
Base area of aquarium: $100 \times 40 = 4000 \, \text{cm}^2$.  
Let $h$ = rise in water level. Volume of displaced water: $4000h = 1000$ → $h=1000/4000=0.25$ cm.  
ANSWER 6: A

---

### Problem 7:
Each principal serves a non-overlapping 3-year term. We want the maximum number of principals serving *at any time* during an 8-year period.  
- A 3-year term only needs to overlap with the 8-year period to count.  
- Example: Let the 8-year period be 2000–2007.  
  - Term 1: 1998–2000 (serves 2000, in period)  
  - Term 2: 2001–2003 (all in period)  
  - Term 3: 2004–2006 (all in period)  
  - Term 4: 2007–2009 (serves 2007, in period)  
This gives 4 principals. 5 terms would require the first term to end before 2000, so it does not count.  
ANSWER 7: C

---

### Problem 8:
Theresa needs an average of 10 hours/week for 6 weeks, so total required hours: $10 \times 6 = 60$.  
Sum of first 5 weeks: $8 + 11 +7 +12 +10 = 48$.  
Hours needed in week 6: $60 - 48 =12$.  
ANSWER 8: D

---

### Problem 9:
Total payment is $1A2 = 100 + 10A + 2 = 102 + 10A$, which is divisible by 11 (11 members paid equal integer amounts).  
Compute modulo 11:  
$102 \mod 11 = 3$, $10A \mod 11 = -A$ (since $10 \equiv -1 \mod 11$).  
So $3 - A \equiv 0 \mod 11$ → $A=3$ (only digit 0-9 that works).  
Check: $132 / 11 = 12$, integer.  
ANSWER 9: D

---

### Problem 10:
July has 31 days, each with 24 hours. Total hours in July: $31 \times 24$.  
Average rainfall per hour = total rain / total hours = $366 / (31 \times 24)$.  
ANSWER 10: A

---

### Problem 11:
Big Al ate bananas for 5 days, 6 more each day, total 100. This is an arithmetic sequence with $n=5$, $d=6$, sum $S=100$.  
Sum formula: $S = \frac{n}{2}[2a + (n-1)d]$ → $100 = \frac{5}{2}[2a + 4*6]$ → $100 =5(a+12)$ → $a=8$ (May 1).  
May 5 is the 5th term: $a +4d =8 +24=32$.  
ANSWER 11: D

---

### Problem 12:
Total age of Room A: $6 \times40=240$. Total age of Room B: $4 \times25=100$.  
Combined total age: $240+100=340$. Total people: $6+4=10$.  
Average age: $340 /10=34$.  
ANSWER 12: D