🔬 Model Lab

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B

Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Compute each group. First group is arithmetic with average 21:
1+11+21+31+41 = 105, since 5 × 21 = 105.
Second group has average 29:
9+19+29+39+49 = 145, since 5 × 29 = 145.

Add: 105 + 145 = 250.

Another way: pair by columns: (1+9)=10, (11+19)=30, (21+29)=50, (31+39)=70, (41+49)=90. Sum = 10+30+50+70+90 = 250.

So 150, 199, 200, 249 are too small from mis-adding.
ANSWER 1: E

Problem 2:
We need the combined average, not just the average of the averages.

Total age in Room A = number × average = 6 × 40 = 240.
Total age in Room B = 4 × 25 = 100.
Combined total = 240 + 100 = 340.
Combined number of people = 6 + 4 = 10.
Combined average = 340 ÷ 10 = 34.

Choice A 32.5 is (40+25)/2, which is wrong because the groups have different sizes. We must weight by 6 and 4.
ANSWER 2: D

Problem 3:
Recall the parts of a cube.

Edges: a cube has 4 around top, 4 around bottom, 4 vertical = 12. Jamie = 12.
Corners, i.e. vertices: 4 on top + 4 on bottom = 8. Jimmy = 8.
Faces: 6 - top, bottom, front, back, left, right. Judy = 6.

Sum = 12 + 8 + 6 = 26.
ANSWER 3: E

Problem 4:
We have an arithmetic sequence decreasing by 7, starting at 100.

1st = 100
2nd = 100 - 7 = 93
3rd = 100 - 2×7 = 86
In general, nth = 100 - 7×(n-1).

For n = 10: subtract 7 nine times:
10th = 100 - 7×9 = 100 - 63 = 37.

Choice A 30 comes from subtracting 7 ten times, which would be the 11th term, an off-by-one error.
ANSWER 4: B

Problem 5:
Factor 250 into primes.

250 = 25 × 10 = 5×5 × 2×5 = 2 × 5 × 5 × 5.

The prime factors present are 2 and 5. The two smallest, counting with multiplicity, are 2 and 5.
Their sum = 2 + 5 = 7.

10 would be 5+5, ignoring the 2, which is smaller.
ANSWER 5: C

Problem 6:
One number is odd from {1,3,5}, the other is even from {2,4,6}. So the sum is always odd.

Smallest sum = 1+2 = 3. Largest sum = 5+6 = 11.
Possible odd integers in between: 3,5,7,9,11.

Check they all occur:
3 = 1+2
5 = 3+2 = 1+4
7 = 5+2 = 3+4 = 1+6
9 = 5+4 = 3+6
11 = 5+6

So there are 5 different sum values.

Choice E 9 counts all 3×3 pairs, but many give the same sum, e.g. 1+6 = 3+4 = 5+2 = 7.
ANSWER 6: B

Problem 7:
Start at noon. Add 1000 minutes.

60 minutes = 1 hour. 1000 ÷ 60 = 16 remainder 40, since 16×60 = 960.
So 1000 min = 16 hours 40 minutes.

Noon + 12 hours = midnight. 4 more hours = 4:00 a.m. next day. Plus 40 minutes = 4:40 a.m.

Check: noon to midnight is 720 min, leaving 280 min = 4 hr 40 min.
ANSWER 7: D

Problem 8:
Find current ages.

Five years ago Bella turned 6. So Bella now = 6+5 = 11.
The kitten was newborn (age 0) five years ago. So kitten now = 0+5 = 5.
Let Anna now = A. Sum today is 30:
A + 11 + 5 = 30
A + 16 = 30
A = 14.

Difference Anna - Bella = 14 - 11 = 3 years.
ANSWER 8: C

Problem 9:
Regular price for 3 pairs = 3 × $50 = $150.

Amount Javier pays:
1st pair: $50 full price.
2nd pair: 40% discount means pay 100%-40% = 60%. 0.60 × 50 = $30.
3rd pair: half price = $50 ÷ 2 = $25.
Total paid = 50 + 30 + 25 = $105.

Amount saved = 150 - 105 = $45.
Percent saved = 45 / 150 = 4500 / 150 %? Compute: 45/150 = 15/50 = 3/10 = 30%.

So he saved 30%, not 40% or 50% which apply only to one pair.
ANSWER 9: B

Problem 10:
Product is a multiple of 6 means divisible by 2 and by 3. So the two dice must supply a factor 2 (an even number) and a factor 3 (a 3 or 6) somewhere.

Test each sum:

Sum 5: 2+3 =5, product 6, multiple of 6. Possible.
Sum 6: pairs are (1,5)=5, (2,4)=8, (3,3)=9, (4,2)=8, (5,1)=5. None is divisible by 6. To be multiple of 6 need even and multiple of 3; no pair has both. So impossible.
Sum 7: 1+6=7, product 6, multiple of 6. Possible. Also 3+4=7 product 12.
Sum 8: 2+6=8, product 12, multiple of 6. Possible.
Sum 9: 3+6=9, product 18, multiple of 6. Possible.

Only 6 cannot occur under the condition.
ANSWER 10: B

Problem 1:

**Restatement of the Problem:**
We are asked to find the value of the sum:
$$(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$$

**Step-by-Step Solution:**
Rather than adding each group separately, we can pair each number from the first group with a corresponding number from the second group to take advantage of the fact that numbers ending in $1$ and $9$ add up to a multiple of $10$:
- $1 + 9 = 10$
- $11 + 19 = 30$
- $21 + 29 = 50$
- $31 + 39 = 70$
- $41 + 49 = 90$

Now, add these five sums together:
$$10 + 30 + 50 + 70 + 90 = 250$$

Alternatively, we could sum each group:
- First group: $1 + 11 + 21 + 31 + 41 = 105$
- Second group: $9 + 19 + 29 + 39 + 49 = 145$
- Total: $105 + 145 = 250$

The sum is $250$, ruling out choices A, B, C, and D.

ANSWER 1: E

---

Problem 2:

**Restatement of the Problem:**
Room A has $6$ people with an average age of $40$. Room B has $4$ people with an average age of $25$. We need to find the average age when all the people from both rooms are combined.

**Step-by-Step Solution:**
1. Calculate the total age of the people in Room A:
   $$\text{Total age in Room A} = 6 \times 40 = 240$$

2. Calculate the total age of the people in Room B:
   $$\text{Total age in Room B} = 4 \times 25 = 100$$

3. Find the combined total age of all people:
   $$\text{Combined age} = 240 + 100 = 340$$

4. Find the total number of people:
   $$\text{Total number of people} = 6 + 4 = 10$$

5. Calculate the combined average age:
   $$\text{Average age} = \frac{340}{10} = 34$$

Note: Simply averaging the two averages, $\frac{40 + 25}{2} = 32.5$, is a common trap (choice A) because the two rooms contain different numbers of people.

ANSWER 2: D

---

Problem 3:

**Restatement of the Problem:**
Jamie counted the edges, Jimmy counted the corners (vertices), and Judy counted the faces of a standard cube. We need to find the sum of these three values.

**Step-by-Step Solution:**
For a standard cube:
- The number of edges is $E = 12$ (4 along the top face, 4 along the bottom face, and 4 vertical edges).
- The number of corners (vertices) is $V = 8$ (4 on top and 4 on the bottom).
- The number of faces is $F = 6$ (top, bottom, and 4 side faces).

Now, add the three counts:
$$\text{Sum} = 12 + 8 + 6 = 26$$

Thus, the sum is $26$.

ANSWER 3: E

---

Problem 4:

**Restatement of the Problem:**
Lucius counts backward by $7$s starting from $100$. The first three numbers in the sequence are $100, 93, 86$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-Step Solution:**
This sequence is an arithmetic progression with first term $a_1 = 100$ and common difference $d = -7$.
The formula for the $n\text{th}$ term is:
$$a_n = a_1 + (n - 1)d$$

For the $10\text{th}$ number ($n = 10$):
$$a_{10} = 100 + (10 - 1)(-7)$$
$$a_{10} = 100 + 9(-7)$$
$$a_{10} = 100 - 63 = 37$$

We can also verify by writing out the terms:
1. $100$
2. $93$
3. $86$
4. $79$
5. $72$
6. $65$
7. $58$
8. $51$
9. $44$
10. $37$

The $10\text{th}$ number is $37$.

ANSWER 4: B

---

Problem 5:

**Restatement of the Problem:**
We are asked to find the sum of the two smallest prime factors of $250$.

**Step-by-Step Solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$

2. Identify the prime factors:
   The prime factors of $250$ are $2$ and $5$.

3. Since $250$ only has two distinct prime factors, the two smallest prime factors are simply $2$ and $5$.

4. Find their sum:
   $$2 + 5 = 7$$

ANSWER 5: C

---

Problem 6:

**Restatement of the Problem:**
Bag A contains chips $\{1, 3, 5\}$ and Bag B contains chips $\{2, 4, 6\}$. One chip is drawn from each bag. We must determine the number of distinct possible values for the sum of the two numbers.

**Step-by-Step Solution:**
Let $a \in \{1, 3, 5\}$ be the number from Bag A, and $b \in \{2, 4, 6\}$ be the number from Bag B.
Let's compute all possible sums $a + b$:
- For $a = 1$:
  - $1 + 2 = 3$
  - $1 + 4 = 5$
  - $1 + 6 = 7$
- For $a = 3$:
  - $3 + 2 = 5$
  - $3 + 4 = 7$
  - $3 + 6 = 9$
- For $a = 5$:
  - $5 + 2 = 7$
  - $5 + 4 = 9$
  - $5 + 6 = 11$

Now, list all the unique values obtained:
$$\{3, 5, 7, 9, 11\}$$

There are $5$ distinct values for the sum.

ANSWER 6: B

---

Problem 7:

**Restatement of the Problem:**
A contest started at 12:00 p.m. (noon) and lasted for $1000$ minutes. We need to find the time the contest ended.

**Step-by-Step Solution:**
1. Convert $1000$ minutes into hours and minutes:
   $$1000 \div 60 = 16 \text{ with a remainder of } 40$$
   So, $1000\text{ minutes} = 16\text{ hours and } 40\text{ minutes}$.

2. Add $16$ hours to 12:00 p.m. (noon):
   - $12$ hours after noon is 12:00 a.m. (midnight).
   - $4$ more hours after midnight is 4:00 a.m.
   So, 12:00 p.m. + $16$ hours = 4:00 a.m. the next day.

3. Add the remaining $40$ minutes:
   $$\text{4:00 a.m.} + 40\text{ minutes} = \text{4:40 a.m.}$$

Thus, the contest ended at 4:40 a.m.

ANSWER 7: D

---

Problem 8:

**Restatement of the Problem:**
Five years ago, Bella was $6$ years old and got a newborn kitten (age $0$). Today, the sum of the ages of Anna, Bella, and the kitten is $30$. We need to find how many years older than Bella Anna is.

**Step-by-Step Solution:**
1. Determine Bella's age today:
   $$\text{Bella's age} = 6 + 5 = 11\text{ years}$$

2. Determine the kitten's age today:
   The kitten was newborn ($0$ years old) $5$ years ago, so:
   $$\text{Kitten's age} = 0 + 5 = 5\text{ years}$$

3. Find Anna's age today using the total sum of $30$:
   $$\text{Anna's age} + \text{Bella's age} + \text{Kitten's age} = 30$$
   $$\text{Anna's age} + 11 + 5 = 30$$
   $$\text{Anna's age} + 16 = 30$$
   $$\text{Anna's age} = 14\text{ years}$$

4. Determine how many years older than Bella Anna is:
   $$\text{Anna's age} - \text{Bella's age} = 14 - 11 = 3\text{ years}$$

ANSWER 8: C

---

Problem 9:

**Restatement of the Problem:**
The regular price of one pair of sandals is $\$50$. The fair special gives:
- First pair at regular price ($\$50$)

Problem 1:
We need to compute (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

Direct approach:
- First group: 1 + 11 = 12; 12 + 21 = 33; 33 + 31 = 64; 64 + 41 = 105.
- Second group: 9 + 19 = 28; 28 + 29 = 57; 57 + 39 = 96; 96 + 49 = 145.
- Total: 105 + 145 = 250.

A nice check: pairing numbers across the groups gives 1+49 = 50, 11+39 = 50, 21+29 = 50, 31+19 = 50, 41+9 = 50, i.e., five pairs of 50 = 250. This confirms the total is 250 (and rules out the near-miss 249, which would come from an arithmetic slip).

ANSWER 1: E

Problem 2:
Averages must be converted to totals before combining (a common trap is to average 40 and 25 to get 32.5 — that's wrong because the group sizes differ).

- Room A: 6 people × 40 = 240 total years.
- Room B: 4 people × 25 = 100 total years.
- Combined: 240 + 100 = 340 total years for 6 + 4 = 10 people.
- Average = 340 ÷ 10 = 34.

ANSWER 2: D

Problem 3:
For a cube: Edges = 12, Corners (vertices) = 8, Faces = 6.
Sum = 12 + 8 + 6 = 26.

(One can verify: a cube is two square pyramids base-to-base, or just picture a box: 4 edges on top, 4 on bottom, 4 vertical = 12 edges; 8 corners; 6 faces.)

ANSWER 3: E

Problem 4:
Lucius's numbers form an arithmetic sequence starting at 100 with common difference −7:
100, 93, 86, 79, 72, 65, 58, 51, 44, 37, ...

Using the formula: n-th term = 100 − 7(n − 1). For n = 10:
100 − 7(9) = 100 − 63 = 37.

Counting the list: 100 (1st), 93 (2nd), 86 (3rd), 79 (4th), 72 (5th), 65 (6th), 58 (7th), 51 (8th), 44 (9th), 37 (10th). Confirmed.

ANSWER 4: B

Problem 5:
Factor 250: 250 = 25 × 10 = 2 × 5³.
The prime factors are 2 and 5 (5 appears three times, but it's still just the prime 5).
The two smallest prime factors are 2 and 5, and their sum is 2 + 5 = 7.

(The tempting wrong answers: 2 alone is only one factor; 5 is the larger of the two; 10 = 2 × 5 is the product, not the sum.)

ANSWER 5: C

Problem 6:
Bag A gives {1, 3, 5} and Bag B gives {2, 4, 6}. List all 9 sums:
- 1+2 = 3, 1+4 = 5, 1+6 = 7
- 3+2 = 5, 3+4 = 7, 3+6 = 9
- 5+2 = 7, 5+4 = 9, 5+6 = 11

Distinct sums: {3, 5, 7, 9, 11}, which is 5 different values.
(Trap: 9 sums occur, but only 5 distinct values, so choice E, 9, is wrong; since all numbers are odd + even = odd, the sums are 3, 5, 7, 9, 11 exactly.)

ANSWER 6: B

Problem 7:
Start at noon (12:00 p.m.) and add 1000 minutes.
Convert: 1000 minutes = 16 hours and 40 minutes (since 16 × 60 = 960, and 1000 − 960 = 40).
- Noon + 16 hours = 4:00 a.m. the next day.
- 4:00 a.m. + 40 minutes = 4:40 a.m.

(Check the near-misses: 10:00 p.m. would be only 10 hours; midnight is 12 hours; 2:30 a.m. is 14.5 hours; 6:40 a.m. is 18 hours 40 minutes — 17 hours 40 minutes too many in total would be needed otherwise. Only 4:40 a.m. matches 16 h 40 min.)

ANSWER 7: D

Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11 years old.
The kitten was newborn (age 0) five years ago, so today the kitten is 5 years old.
Today the sum of Anna's, Bella's, and the kitten's ages is 30:
Anna + 11 + 5 = 30, so Anna = 30 − 16 = 14 years old.
Anna is older than Bella by 14 − 11 = 3 years.

(Trap: don't try to average or subtract the kitten's age from 30 incorrectly — the equation above handles everyone directly.)

ANSWER 8: C

Problem 9:
Regular price per pair: $50. Javier buys three pairs.
Without the special, three pairs would cost 3 × $50 = $150.
With the special:
- 1st pair: $50
- 2nd pair: 40% off, so $50 − 0.40 × $50 = $50 − $20 = $30
- 3rd pair: half price = $25
Total paid: 50 + 30 + 25 = $105.

Savings: $150 − $105 = $45.
Percentage saved: 45/150 = 0.30 = 30%.

(Check the traps: 40% is the discount on only the second pair, not the overall savings; the overall savings is 45/150 = 30%, not 45%.)

ANSWER 9: B

Problem 10:
The product of the two dice must be a multiple of 6 (divisible by both 2 and 3). Check each proposed sum by listing the rolls that produce it:

- Sum 5: rolls (2,3) or (3,2) give product 6. ✓ Possible.
- Sum 6: rolls are (1,5), (2,4), (3,3), (4,2), (5,1). Products: 5, 8, 9, 8, 5. None is divisible by 6. ✗ Impossible.
- Sum 7: roll (1,6) gives product 6, or (3,4) gives product 12. ✓ Possible.
- Sum 8: roll (2,6) gives product 12. ✓ Possible.
- Sum 9: roll (3,6) gives product 18. ✓ Possible.

Since sum 6 has no roll whose product is a multiple of 6, it cannot occur.

ANSWER 10: B

Problem 1:  
We need the sum of two groups:

\[
(1+11+21+31+41)+(9+19+29+39+49)
\]

First group:

\[
1+11+21+31+41=105
\]

Second group:

\[
9+19+29+39+49=145
\]

Total:

\[
105+145=250
\]

A tempting mistake would be to only add one group or mispair the numbers, but the full sum is 250.

ANSWER 1: E

Problem 2:  
The average age of the 6 people in Room A is 40, so their total age is

\[
6\cdot 40=240
\]

The average age of the 4 people in Room B is 25, so their total age is

\[
4\cdot 25=100
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340
\]

The combined average age is

\[
\frac{340}{10}=34
\]

A tempting wrong choice is to average the two averages:

\[
\frac{40+25}{2}=32.5
\]

but that ignores that Room A has more people. The correct weighted average is 34.

ANSWER 2: D

Problem 3:  
A cube has:

- 12 edges
- 8 corners, also called vertices
- 6 faces

The problem says Jamie counted the edges, Jimmy counted the corners, and Judy counted the faces, then they added the three numbers.

So the sum is

\[
12+8+6=26
\]

ANSWER 3: E

Problem 4:  
Lucius is counting backward by 7s. His first number is 100.

So:

- 1st number: \(100\)
- 2nd number: \(100-7=93\)
- 3rd number: \(93-7=86\)

The \(n\)th number is

\[
100-7(n-1)
\]

For the 10th number:

\[
100-7(10-1)=100-7\cdot 9=100-63=37
\]

A common mistake is to subtract \(7\cdot 10=70\), which would give 30, but that would be the 11th number, not the 10th.

ANSWER 4: B

Problem 5:  
We need the sum of the two smallest prime factors of 250.

Factor 250:

\[
250=2\cdot 125=2\cdot 5^3
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

Their sum is

\[
2+5=7
\]

ANSWER 5: C

Problem 6:  
Bag A has chips labeled \(1,3,5\). Bag B has chips labeled \(2,4,6\). One chip is drawn from each bag.

All possible sums are:

\[
\begin{aligned}
1+2&=3\\
1+4&=5\\
1+6&=7\\
3+2&=5\\
3+4&=7\\
3+6&=9\\
5+2&=7\\
5+4&=9\\
5+6&=11
\end{aligned}
\]

The different values are

\[
3,5,7,9,11
\]

There are 5 different possible sums.

A tempting wrong answer is 9, because there are 3 chips in each bag, giving \(3\cdot 3=9\) outcomes, but the question asks for different sums, not different outcomes.

ANSWER 6: B

Problem 7:  
The contest began at noon and lasted 1000 minutes.

Convert 1000 minutes to hours:

\[
1000\div 60=16 \text{ hours } 40 \text{ minutes}
\]

Starting at noon:

\[
12:00\text{ p.m.}+16\text{ hours}=4:00\text{ a.m.}
\]

Then add 40 minutes:

\[
4:00\text{ a.m.}+40\text{ minutes}=4:40\text{ a.m.}
\]

So the contest ended at 4:40 a.m.

ANSWER 7: D

Problem 8:  
Five years ago, Bella was 6. So today, Bella is

\[
6+5=11
\]

Five years ago, the kitten was newborn, so it is now 5 years old.

Let Anna’s current age be \(a\). The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
a+11+5=30
\]

\[
a+16=30
\]

\[
a=14
\]

Anna is 14 and Bella is 11, so Anna is

\[
14-11=3
\]

years older than Bella.

ANSWER 8: C

Problem 9:  
The regular price of three pairs of sandals is

\[
3\cdot 50=150
\]

Under the special:

- First pair: full price, \(50\)
- Second pair: 40% off, so he pays 60% of 50:

\[
0.60\cdot 50=30
\]

- Third pair: half price:

\[
\frac{50}{2}=25
\]

Total paid:

\[
50+30+25=105
\]

Amount saved:

\[
150-105=45
\]

The percentage of the $150 regular price that he saved is

\[
\frac{45}{150}\cdot 100\%=30\%
\]

A tempting wrong answer is 40%, but that is only the discount on the second pair, not the total discount on all three pairs.

ANSWER 9: B

Problem 10:  
Aaliyah rolls two standard dice. The product is a multiple of 6. We need to determine which of the given sums cannot occur.

A product of two numbers is a multiple of 6 if it is divisible by both 2 and 3. With dice numbers from 1 to 6, this happens when:

- one die shows 3 or 6, and
- at least one die is even.

Let’s list possible sums that work:

- Product \(2\cdot 3=6\), sum \(2+3=5\)
- Product \(3\cdot 4=12\), sum \(3+4=7\)
- Product \(2\cdot 6=12\), sum \(2+6=8\)
- Product \(3\cdot 6=18\), sum \(3+6=9\)

So 5, 7, 8, and 9 are possible.

Now check whether sum 6 is possible. The pairs of dice with sum 6 are:

\[
(1,5),\ (2,4),\ (3,3)
\]

Their products are:

\[
1\cdot 5=5,\quad 2\cdot 4=8,\quad 3\cdot 3=9
\]

None of these products is a multiple of 6. Therefore, 6 cannot be the sum.

ANSWER 10: B

# Problem 1:
We need the value of (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).

First group: 1 + 11 = 12, +21 = 33, +31 = 64, +41 = **105**.
Second group: 9 + 19 = 28, +29 = 57, +39 = 96, +49 = **145**.

Total: 105 + 145 = **250**.

(A quick check: pair terms 1+9=10, 11+19=30, 21+29=50, 31+39=70, 41+49=90, and 10+30+50+70+90 = 250.) The choice 250 is right; 249 or 200 come from arithmetic slips.

ANSWER 1: E

# Problem 2:
Average age must be computed from *total* ages, not by averaging the averages.

Room A total: 6 × 40 = 240.
Room B total: 4 × 25 = 100.
Combined total: 240 + 100 = 340 for 6 + 4 = 10 people.

Average = 340 / 10 = **34**.

(The trap answer 32.5 is the average of 40 and 25, which ignores that Room A has more people, pulling the average up toward 40.)

ANSWER 2: D

# Problem 3:
A cube has:
- edges: 12
- corners (vertices): 8
- faces: 6

Sum = 12 + 8 + 6 = **26**.

(Euler check: V − E + F = 8 − 12 + 6 = 2 ✓.)

ANSWER 3: E

# Problem 4:
The sequence is 100, 93, 86, … decreasing by 7. The nth number is 100 − 7(n − 1).

For n = 10: 100 − 7(9) = 100 − 63 = **37**.

(Answer 44 comes from subtracting only 8 sevens — i.e., miscounting which term is 10th.)

ANSWER 4: B

# Problem 5:
Factor 250: 250 = 2 × 125 = 2 × 5³.

So the prime factors are 2 and 5 only. The two smallest are 2 and 5.

Sum = 2 + 5 = **7**.

(Answers 10 or 12 would require primes like 5+5 or other factors that aren't there.)

ANSWER 5: C

# Problem 6:
List all 3 × 3 = 9 sums (one chip from each bag):

- 1: 1+2=3, 1+4=5, 1+6=7
- 3: 3+2=5, 3+4=7, 3+6=9
- 5: 5+2=7, 5+4=9, 5+6=11

Distinct sums: {3, 5, 7, 9, 11} → **5 values**.

(9 counts all outcomes, not distinct sums; all sums are odd, so only these five odd numbers from 3 to 11 occur.)

ANSWER 6: B

# Problem 7:
Convert 1000 minutes to hours: 1000 ÷ 60 = 16 hours with remainder 40 minutes.

Starting at 12:00 noon: +16 hours = 4:00 a.m. the next day; +40 minutes = **4:40 a.m.**

(10:00 p.m. would be 600 minutes; midnight is 720 minutes.)

ANSWER 7: D

# Problem 8:
Five years ago Bella turned 6, so today Bella is 6 + 5 = 11.
The kitten was newborn (age 0) then, so today the kitten is 5.

Sum of all three ages today = 30, so Anna's age = 30 − 11 − 5 = 14.

Anna is 14 − 11 = **3 years older** than Bella.

ANSWER 8: C

# Problem 9:
Regular price of three pairs: 3 × $50 = $150.

What Javier paid:
- 1st pair: $50
- 2nd pair: 40% off → 0.60 × 50 = $30
- 3rd pair: half price → $25

Total paid: 50 + 30 + 25 = $105.
Savings: 150 − 105 = $45.

Percent saved: 45/150 = 0.30 = **30%**.

(45% is the dollar amount saved, not the percentage; 25%+... type averaging also misleads.)

ANSWER 9: B

# Problem 10:
For a product to be a multiple of 6, it needs a factor of 2 and a factor of 3. Let's find all such dice pairs and their sums:

- Any pair containing a 6: (1,6)=7, (2,6)=8, (3,6)=9, (4,6)=10, (5,6)=11, (6,6)=12
- Pairs with a 3 and an even number: (3,2)=5, (3,4)=7, (3,6)=9

Possible sums: 5, 7, 8, 9, 10, 11, 12.

Now check sum 6: the pairs are (1,5) product 5, (2,4) product 8, (3,3) product 9 — none is a multiple of 6.

So the sum that **cannot** occur is **6**.

ANSWER 10: B

Problem 1:

We need to add two groups of five numbers:

\[
(1+11+21+31+41)+(9+19+29+39+49).
\]

Pair corresponding terms:

\[
1+9=10,\quad 11+19=30,\quad 21+29=50,
\]
\[
31+39=70,\quad 41+49=90.
\]

Now add:

\[
10+30+50+70+90=250.
\]

Choices A through D are too small. The correct choice is E.

ANSWER 1: E

Problem 2:

Room A has 6 people with average age 40, so their total age is

\[
6\cdot 40=240.
\]

Room B has 4 people with average age 25, so their total age is

\[
4\cdot 25=100.
\]

Together, there are

\[
6+4=10
\]

people, with total age

\[
240+100=340.
\]

Therefore, the combined average is

\[
\frac{340}{10}=34.
\]

A simple average of 40 and 25 would be incorrect because the rooms have different numbers of people. The correct choice is D.

ANSWER 2: D

Problem 3:

A cube has:

- 12 edges,
- 8 corners (vertices),
- 6 faces.

Their sum is

\[
12+8+6=26.
\]

Choices A through D leave out or miscount one of these cube features. The correct choice is E.

ANSWER 3: E

Problem 4:

Lucius counts backward by 7 each time:

\[
100,\ 93,\ 86,\dots
\]

The first number is 100. To get to the 10th number, we subtract 7 a total of 9 times:

\[
100-9(7)=100-63=37.
\]

Indeed, the list is

\[
100,93,86,79,72,65,58,51,44,37.
\]

The 10th number is 37, which is choice B.

ANSWER 4: B

Problem 5:

Factor 250:

\[
250=2\cdot125=2\cdot5\cdot5\cdot5.
\]

The prime factors are 2 and 5. The two smallest prime factors are therefore 2 and 5.

\[
2+5=7.
\]

Choice A counts only 2, while choices D and E incorrectly add repeated factors. The correct choice is C.

ANSWER 5: C

Problem 6:

Bag A contains odd numbers:

\[
1,3,5.
\]

Bag B contains even numbers:

\[
2,4,6.
\]

Every sum will be odd. List the possible sums:

\[
1+2=3,\quad 1+4=5,\quad 1+6=7,
\]

\[
3+2=5,\quad 3+4=7,\quad 3+6=9,
\]

\[
5+2=7,\quad 5+4=9,\quad 5+6=11.
\]

The different possible sums are

\[
3,5,7,9,11.
\]

There are 5 different values. Choice E, 9, would count all possible chip pairs rather than distinct sums. The correct choice is B.

ANSWER 6: B

Problem 7:

There are 60 minutes in an hour.

\[
1000\div 60=16\text{ hours with }40\text{ minutes left}.
\]

Starting at noon and adding 12 hours gives midnight. Then add the remaining 4 hours 40 minutes:

\[
\text{midnight}+4\text{ hours }40\text{ minutes}=4{:}40\text{ a.m.}
\]

Thus the contest ended at 4:40 a.m., choice D.

ANSWER 7: D

Problem 8:

Five years ago, Bella turned 6. Therefore today Bella is

\[
6+5=11
\]

years old.

The kitten was newborn five years ago, so today the kitten is

\[
5
\]

years old.

The sum of Anna’s age, Bella’s age, and the kitten’s age is 30:

\[
\text{Anna}+11+5=30.
\]

So Anna’s age is

\[
30-16=14.
\]

Anna is therefore older than Bella by

\[
14-11=3
\]

years.

The correct choice is C.

ANSWER 8: C

Problem 9:

The regular total price for three pairs is

\[
3\cdot \$50=\$150.
\]

Javier pays:

- First pair: \(\$50\)
- Second pair: 40% discount means he pays 60% of \(\$50\):

\[
0.60(50)=\$30
\]

- Third pair: half price:

\[
\frac12(50)=\$25
\]

Total paid:

\[
50+30+25=\$105.
\]

His savings are

\[
150-105=\$45.
\]

The percentage saved is

\[
\frac{45}{150}=0.30=30\%.
\]

The correct choice is B.

ANSWER 9: B

Problem 10:

The product must be a multiple of 6. Since

\[
6=2\cdot 3,
\]

the two dice must together provide a factor of 2 and a factor of 3.

Check the possible sums in the choices:

- Sum 5: \(2+3=5\), and \(2\cdot3=6\), so 5 is possible.
- Sum 6: possibilities are \(1+5,\ 2+4,\ 3+3\). Their products are

\[
1\cdot5=5,\quad 2\cdot4=8,\quad 3\cdot3=9,
\]

none of which is divisible by 6. So 6 is impossible.
- Sum 7: \(1+6=7\), and \(1\cdot6=6\), so 7 is possible.
- Sum 8: \(2+6=8\), and \(2\cdot6=12\), so 8 is possible.
- Sum 9: \(3+6=9\), and \(3\cdot6=18\), so 9 is possible.

Therefore, the sum that cannot occur is 6, choice B.

ANSWER 10: B