🔬 Model Lab

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro

Problem 1:
We are asked for the number of 4-digit strings using digits 0-9 with repetition allowed, except we forbid those beginning with 9,1,1.

Total without restriction: each of the 4 positions has 10 choices.
So total = 10 × 10 × 10 × 10 = 10,000.

"Begin with the sequence 9,1,1" means the first three digits are 9, 1, 1 in order. The fourth digit can be anything 0-9.
So forbidden passwords are of form 9-1-1-?:
9110, 9111, ..., 9119.
That is 1 × 1 × 1 × 10 = 10 passwords.

Hence allowed = 10,000 - 10 = 9,990.

This is not 9,999, which would subtract only one password, and not 9,000 which would forbid a whole first digit.
ANSWER 1: D

Problem 2:
We need the mean weight of all 10 children together.

Total weight of 6 boys = 6 × average = 6 × 150 = 900 pounds.
Total weight of 4 girls = 4 × average = 4 × 120 = 480 pounds.
Combined total = 900 + 480 = 1,380 pounds.
Number of children = 6 + 4 = 10.

Average = total / number = 1380 / 10 = 138 pounds.

Note this is not the simple average (150+120)/2 = 135, because there are more boys than girls, so the average is weighted toward 150.
ANSWER 2: C

Problem 3:
Bert's box capacity is 125 jellybeans. Carrie's box is twice as high, twice as wide, twice as long.

Volume scales as length × width × height.
If each linear dimension is multiplied by 2, volume is multiplied by:
2 × 2 × 2 = 2³ = 8.

Assuming jellybeans pack similarly, number scales with volume:
Carrie's number ≈ 125 × 8 = 1,000.

Doubling each side does not double volume; that would give 250. Tripling volume ideas give 750, etc., are wrong.
ANSWER 3: E

Problem 4:
We need the decimal expansion of 4/37 and its 100th digit after the point.

Note 37 × 27 = 999. So:
1/37 = 27 / (37×27) = 27/999 = 0.027027027...
Therefore:
4/37 = 4 × 1/37 = 108/999 = 0.108108108...

So the decimal repeats with period 3: "108" over and over.
Position 1: 1, position 2: 0, position 3: 8, position 4: 1, etc.

For position n, the digit depends on n mod 3:
n ≡ 1 mod 3 → 1
n ≡ 2 mod 3 → 0
n ≡ 0 mod 3 → 8

100 = 99 + 1, so 100 ≡ 1 mod 3. So the 100th digit is the same as the 1st: 1.
ANSWER 4: B

Problem 5:
Ratio boys : girls = 2 : 3. Total students = 30.

Total parts = 2 + 3 = 5 parts.
Value of one part = 30 ÷ 5 = 6 students.

Boys = 2 parts = 2 × 6 = 12.
Girls = 3 parts = 3 × 6 = 18.

More girls than boys = 18 - 12 = 6.

It is not 10, which is the number in each... etc.; compute carefully.
ANSWER 5: D

Problem 6:
Gage skates 5 days at 1 hr 15 min, 3 days at 1 hr 30 min, plus a 9th day. Want 9-day average 85 min.

Convert: 1 hr 15 min = 75 min. 1 hr 30 min = 90 min.

Already skated:
5 × 75 = 375 min.
3 × 90 = 270 min.
Sum so far for 8 days = 375 + 270 = 645 min.

Want for 9 days average 85:
required total = 9 × 85 = 765 min.

Need on day 9:
765 - 645 = 120 min = 2 hr.

Shorter times like 1 hr would leave the average below 85.
ANSWER 6: E

Problem 7:
Count whole numbers between 100 and 400 that contain the digit 2. 100 and 400 themselves do not contain 2, so inclusive/exclusive does not matter.

Split by hundreds:

200-299: hundreds digit is 2, so every number contains 2.
That is 100 numbers.

100-199: need tens =2 or ones=2.
Tens=2: 120-129 → 10 numbers.
Ones=2: 102,112,...,192 → 10 numbers.
Overlap both: 122 counted twice.
So 10+10-1 = 19 numbers.

300-399: same logic as 100-199.
Tens=2: 320-329 → 10.
Ones=2: 302,...,392 → 10.
Overlap: 322.
So 19 numbers.

Total = 100 + 19 + 19 = 138.

Forgetting to subtract the double-counted 122 and 322 would give 140, which is wrong.
ANSWER 7: C

Problem 8:
Square side 10. Lemming ends up somewhere inside the square. Let its distances to the four sides be a,b,c,d, with a,c opposite and b,d opposite.

For any interior point, sum of distances to a pair of opposite sides equals the side length:
a + c = 10,
b + d = 10.
So a+b+c+d = 20.
Average of the four = 20/4 = 5.

We should check the lemming is still inside, so this applies. Starting at a corner, go 6.2 along diagonal (diagonal length is 10√2 ≈14.1, so 6.2 stays inside), then 2 m perpendicular still leaves it inside a 10×10 square. So the formula holds.

The numbers 6.2 and 2 are irrelevant distractors; the answer is always 5 for any interior point.
ANSWER 8: C

Problem 9:
The 25 numbers form a 5×5 array:

Row1: 10  6  4  3  2
Row2: 11  7 14 10  8
Row3:  8  3  4  5  9
Row4: 13  4 15 12  1
Row5:  8  2  5  9  3

Find row minima:
Row1 min = 2 (col5). Col5 = 2,8,9,1,3, max=9 ≠2.
Row2 min = 7 (col2). Col2 = 6,7,3,4,2, max=7. This matches!
Row3 min = 3 (col2). Col2 max=7 ≠3.
Row4 min = 1 (col5). Col5 max=9 ≠1.
Row5 min = 2 (col2). Col2 max=7 ≠2.

Only 7 is smallest in its row (Row2: 11,7,14,10,8) and largest in its column (Col2: 6,7,3,4,2).

No other number satisfies both.
ANSWER 9: C

Problem 10:
Let first four equal scores be x, last score be y. Integers 0 to 100, x < y.

Average 82 over 5 tests:
(4x + y)/5 = 82
4x + y = 410
y = 410 - 4x.

Constraints:
y ≤ 100 → 410 - 4x ≤ 100 → 4x ≥ 310 → x ≥ 77.5 → x ≥ 78.
y > x → 410 - 4x > x → 410 > 5x → x < 82 → x ≤ 81.
Also 0 ≤ x ≤ 100 automatically satisfied.

So x can be 78, 79, 80, 81.
Corresponding y:
78 → 410-312=98
79 → 410-316=94
80 → 410-320=90
81 → 410-324=86
All are ≤100 and >x and integers.

Thus there are 4 distinct possible values for the last test: 86,90,94,98.
ANSWER 10: A

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement and Setup:**
We need to find the number of 4-digit ATM passwords using digits from 0 to 9, where repeats are allowed, but the password cannot begin with the sequence $9, 1, 1$.

**Computation:**
1. A 4-digit password has 4 positions. Each position can be filled by any of the 10 digits from $0$ to $9$.
   $$\text{Total number of passwords} = 10 \times 10 \times 10 \times 10 = 10{,}000$$
2. The restricted passwords begin with the sequence $9, 1, 1$. These take the form $9, 1, 1, d$, where $d$ is any digit from $0$ to $9$.
   $$\text{Number of restricted passwords} = 10 \text{ (from } 9110 \text{ to } 9119\text{)}$$
3. Subtract the restricted passwords from the total:
   $$10{,}000 - 10 = 9{,}990$$

ANSWER 1: D

---

### Problem 2:
**Restatement and Setup:**
We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Computation:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900 \text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480 \text{ pounds}$$
3. Find the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380 \text{ pounds}$$
4. Calculate the overall average:
   $$\text{Average weight} = \frac{1380}{10} = 138 \text{ pounds}$$

ANSWER 2: C

---

### Problem 3:
**Restatement and Setup:**
Bert's box holds 125 jellybeans. Carrie's box is twice as high, twice as wide, and twice as long as Bert's. We need to find approximately how many jellybeans Carrie's box can hold.

**Computation:**
1. The capacity of a box is proportional to its volume:
   $$\text{Volume} = \text{length} \times \text{width} \times \text{height}$$
2. If all three dimensions are doubled, the volume increases by a factor of:
   $$2 \times 2 \times 2 = 2^3 = 8$$
3. Therefore, Carrie's box holds 8 times as many jellybeans as Bert's box:
   $$125 \times 8 = 1000 \text{ jellybeans}$$

ANSWER 3: E

---

### Problem 4:
**Restatement and Setup:**
We want to find the 100th digit to the right of the decimal point in the decimal expansion of $\frac{4}{37}$.

**Computation:**
1. Convert $\frac{4}{37}$ to its decimal representation:
   $$\frac{1}{37} = 0.027027\ldots = 0.\overline{027}$$
   $$\frac{4}{37} = 4 \times 0.027027\ldots = 0.108108\ldots = 0.\overline{108}$$
2. The repeating block is "108", which has a period length of 3:
   - 1st digit: 1
   - 2nd digit: 0
   - 3rd digit: 8
3. Divide the position 100 by 3 to find its position in the repeating cycle:
   $$100 = 3 \times 33 + 1$$
   The remainder is 1, which means the 100th digit is the 1st digit of the repeating block, which is $1$.

ANSWER 4: B

---

### Problem 5:
**Restatement and Setup:**
The ratio of boys to girls is $2 : 3$, and there are 30 students in total. We need to find how many more girls than boys are in the class.

**Computation:**
1. The ratio has $2 + 3 = 5$ equal parts.
2. The number of students per part is:
   $$\frac{30}{5} = 6$$
3. Number of boys:
   $$2 \times 6 = 12$$
4. Number of girls:
   $$3 \times 6 = 18$$
5. The difference between the number of girls and boys is:
   $$18 - 12 = 6$$

ANSWER 5: D

---

### Problem 6:
**Restatement and Setup:**
Gage skated:
- 1 hr 15 min each day for 5 days
- 1 hr 30 min each day for 3 days
We need to determine how long he must skate on the 9th day to average 85 minutes per day over all 9 days.

**Computation:**
1. Convert times to minutes:
   - $1\text{ hr } 15\text{ min} = 75\text{ minutes}$
   - $1\text{ hr } 30\text{ min} = 90\text{ minutes}$
2. Calculate the total time skated in the first 8 days:
   $$\text{First 5 days: } 5 \times 75 = 375\text{ minutes}$$
   $$\text{Next 3 days: } 3 \times 90 = 270\text{ minutes}$$
   $$\text{Total so far} = 375 + 270 = 645\text{ minutes}$$
3. Calculate the total time needed for an 85-minute average over 9 days:
   $$9 \times 85 = 765\text{ minutes}$$
4. Calculate the time needed on the 9th day:
   $$765 - 645 = 120\text{ minutes} = 2\text{ hours}$$

ANSWER 6: E

---

### Problem 7:
**Restatement and Setup:**
We need to find the number of integers strictly between 100 and 400 (i.e., from 101 to 399 inclusive) that contain the digit 2.

**Computation:**
Break the range into hundreds:
1. **From 101 to 199:**
   - Numbers with tens digit 2: $120, 121, \ldots, 129$ (10 numbers)
   - Numbers with units digit 2 and tens digit not 2: $102, 112, 132, 142, 152, 162, 172, 182, 192$ (9 numbers)
   - Total in this range: $10 + 9 = 19$ numbers.
2. **From 200 to 299:**
   - Every number in this range has a hundreds digit of 2.
   - Total in this range: $100$ numbers.
3. **From 300 to 399:**
   - Following the same pattern as 101 to 199, there are $19$ numbers containing a 2.
4. Total count:
   $$19 + 100 + 19 = 138$$

ANSWER 7: C

---

### Problem 8:
**Restatement and Setup:**
A lemming is inside a square of side length 10 meters. We need to find the average of the shortest distances from the lemming to the four sides of the square.

**Computation:**
1. Set up a coordinate system where the square's corners are at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$.
2. For any point $(x, y)$ inside the square:
   - Distance to the left side: $x$
   - Distance to the right side: $10 - x$
   - Distance to the bottom side: $y$
   - Distance to the top side: $10 - y$
3. Notice that the sum of the horizontal distances is:
   $$x + (10 - x) = 10$$
   And the sum of the vertical distances is:
   $$y + (10 - y) = 10$$
4. Since the diagonal of the square has length $10\sqrt{2} \approx 14.14$ meters, after running $6.2$ meters along the diagonal and turning $90^\circ$ to run $2$ meters, the lemming remains strictly inside the square.
5. The sum of the shortest distances to all four sides is:
   $$10 + 10 = 20\text{ meters}$$
6. The average of these four distances is:
   $$\frac{20}{4} = 5\t

# Problem 1:
A password is four digits (0–9), repeats allowed, so without restrictions there are $10 \times 10 \times 10 \times 10 = 10{,}000$ passwords.

The forbidden passwords are those beginning with 9, 1, 1. That fixes the first three digits, leaving the fourth digit free: $10$ forbidden passwords.

So the count is $10{,}000 - 10 = 9{,}999 - 1 = 9{,}990$.

(Note: 9999 would only subtract one password; 9000 would wrongly ban *all* passwords starting with 9.)

ANSWER 1: D

# Problem 2:
Find the total weights, not the average of averages.

- 6 boys at 150 lb: $6 \times 150 = 900$ pounds
- 4 girls at 120 lb: $4 \times 120 = 480$ pounds
- Total: $900 + 480 = 1380$ pounds for all 10 children

Average $= 1380 \div 10 = 138$ pounds.

(135 is the trap: it's the unweighted average of 150 and 120, ignoring that there are more boys than girls.)

ANSWER 2: C

# Problem 3:
Capacity depends on volume, and volume scales by the product of the scale factors in each dimension.

Carrie's box is twice as high, twice as wide, twice as long, so its volume is
$$2 \times 2 \times 2 = 8$$
times Bert's volume.

Jellybeans: $8 \times 125 = 1000$.

(250 and 500 are traps from multiplying by 2 or 4 instead of 8.)

ANSWER 3: E

# Problem 4:
Compute the decimal expansion of $\frac{4}{37}$ by long division:

- $40 \div 37 = 1$, remainder $3$
- $30 \div 37 = 0$, remainder $30$
- $300 \div 37 = 8$, remainder $4$ → back to the start

So $\frac{4}{37} = 0.\overline{108} = 0.108108108\ldots$, a repeating block of length 3: "1, 0, 8".

Since $100 = 3 \times 33 + 1$, the 100th digit is the **1st** digit of the block, which is **1**.

ANSWER 4: B

# Problem 5:
The ratio 2 : 3 means 5 equal parts total. With 30 students, each part is
$$30 \div 5 = 6 \text{ students}.$$

- Boys: $2 \times 6 = 12$
- Girls: $3 \times 6 = 18$

Difference: $18 - 12 = 6$ more girls than boys.

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Total so far (8 days):
$$75 \times 5 + 90 \times 3 = 375 + 270 = 645 \text{ minutes}.$$

To average 85 min over 9 days, he needs
$$85 \times 9 = 765 \text{ minutes total}.$$

Ninth day: $765 - 645 = 120$ minutes $= 2$ hours.

ANSWER 6: E

# Problem 7:
Count the numbers from 100 to 400 that **contain** a 2, by hundreds digit:

- **200s (200–299):** every one of these 100 numbers contains a 2 in the hundreds place → 100 numbers.
- **100s and 300s:** count the numbers *without* a 2. The tens digit has 9 choices (not 2) and the units digit has 9 choices (not 2), so $9 \times 9 = 81$ avoid the digit 2. Thus $100 - 81 = 19$ numbers in each of these blocks contain a 2.
- **400:** no 2.

Total: $100 + 19 + 19 = 138$.

(Including or excluding the endpoints 100 and 400 doesn't matter — neither contains a 2.)

ANSWER 7: C

# Problem 8:
Put the square in coordinates with corners $(0,0)$ and $(10,10)$; the lemming starts at $(0,0)$.

Key fact: for **any** point $(a,b)$ inside the square, the distances to the four sides are
$$a, \quad b, \quad 10-a, \quad 10-b,$$
and their sum is
$$a + b + (10-a) + (10-b) = 20.$$

(For example, after the diagonal run the lemming is at $\left(\tfrac{6.2}{\sqrt2},\tfrac{6.2}{\sqrt2}\right) \approx (4.38, 4.38)$; after the 2 m right turn it is at about $(5.80, 2.97)$ — still inside — and the four distances $5.80 + 2.97 + 4.20 + 7.03$ still sum to 20.)

So the average distance is always $20 \div 4 = 5$ meters, no matter where the lemming stops.

ANSWER 8: C

# Problem 9:
Arrange the 25 entries into 5 rows of 5:

| Row | Entries | Smallest in row |
|---|---|---|
| 1 | 10, 6, 4, 3, 2 | 2 |
| 2 | 11, 7, 14, 10, 8 | **7** |
| 3 | 8, 3, 4, 5, 9 | 3 |
| 4 | 13, 4, 15, 12, 1 | 1 |
| 5 | 8, 2, 5, 9, 3 | 2 |

Largest in each column:
- Col 1: max(10, 11, 8, 13, 8) = 13
- Col 2: max(6, 7, 3, 4, 2) = **7**
- Col 3: max(4, 14, 4, 15, 5) = 15
- Col 4: max(3, 10, 5, 12, 9) = 12
- Col 5: max(2, 8, 9, 1, 3) = 9

Check which column-maximum is also its row's minimum: 13, 15, and 12 are in row 4 (whose minimum is 1); 9 is in row 5 (whose minimum is 2). But **7** is the largest in column 2 **and** the smallest in row 2. ✓

ANSWER 9: C

# Problem 10:
Let $x$ be the score on each of the first four tests and $y$ the (higher) last-test score. Then:
$$4x + y = 5 \times 82 = 410 \quad\Longrightarrow\quad y = 410 - 4x.$$

Constraints: $x, y$ integers with $0 \le x, y \le 100$ and $y > x$.

- $y \le 100$: $410 - 4x \le 100 \Rightarrow x \ge 77.5 \Rightarrow x \ge 78$.
- $y > x$: $410 - 4x > x \Rightarrow 410 > 5x \Rightarrow x \le 81$.

So $x \in \{78, 79, 80, 81\}$, giving:
- $x=78 \Rightarrow y=98$
- $x=79 \Rightarrow y=94$
- $x=80 \Rightarrow y=90$
- $x=81 \Rightarrow y=86$

(Check the boundary cases fail: $x=77$ gives $y=102 > 100$; $x=82$ gives $y=82$, not higher than $x$.)

There are $4$ possible values for the last test score.

ANSWER 10: A

Problem 1:  
We need a 4-digit ATM password using digits 0–9, with repeats allowed, except it cannot begin with the three digits 9, 1, 1.

Total possible 4-digit strings:  
\[
10^4=10000
\]

The forbidden passwords have the form  
\[
9110,9111,\ldots,9119
\]
There are 10 of them.

So the number of allowed passwords is  
\[
10000-10=9990
\]

This rules out 9999, which would occur if only one password were forbidden, and 9000, which would occur if leading 0 were not allowed.

ANSWER 1: D

Problem 2:  
The average of the boys’ weights is 150 pounds for 6 boys, so their total weight is  
\[
6\cdot150=900
\]

The average of the girls’ weights is 120 pounds for 4 girls, so their total weight is  
\[
4\cdot120=480
\]

Total weight of all 10 children:  
\[
900+480=1380
\]

Average weight:  
\[
\frac{1380}{10}=138
\]

The tempting choice 135 is just the average of 150 and 120, but that ignores that there are more boys than girls.

ANSWER 2: C

Problem 3:  
Bert’s box holds 125 jellybeans. Carrie’s box is twice as high, twice as wide, and twice as long.

When all three dimensions are doubled, the volume is multiplied by  
\[
2\cdot2\cdot2=8
\]

So Carrie’s box holds approximately  
\[
8\cdot125=1000
\]
jellybeans.

ANSWER 3: E

Problem 4:  
We need the 100th digit after the decimal point of  
\[
\frac4{37}
\]

Divide:
\[
\frac4{37}=0.108108108\ldots
\]

The repeating block is \(108\), with period 3.

Since  
\[
100\equiv 1 \pmod 3,
\]
the 100th digit is the first digit of the block \(108\), which is 1.

ANSWER 4: B

Problem 5:  
The ratio of boys to girls is \(2:3\), so the total number of “parts” is  
\[
2+3=5
\]

There are 30 students, so each part represents  
\[
\frac{30}{5}=6
\]
students.

Number of boys:  
\[
2\cdot6=12
\]

Number of girls:  
\[
3\cdot6=18
\]

Difference:  
\[
18-12=6
\]

ANSWER 5: D

Problem 6:  
Gage skated for 8 days so far and needs the 9th day’s time so that his average over 9 days is 85 minutes.

Convert skating times to minutes:

- 1 hr 15 min = 75 min
- 1 hr 30 min = 90 min

For 5 days:
\[
5\cdot75=375
\]

For 3 days:
\[
3\cdot90=270
\]

Total so far:
\[
375+270=645
\]

For an 85-minute average over 9 days, total needed is
\[
9\cdot85=765
\]

So the ninth day must be
\[
765-645=120
\]
minutes, which is 2 hours.

ANSWER 6: E

Problem 7:  
We need whole numbers between 100 and 400 that contain the digit 2. Including or excluding 100 and 400 does not matter because neither contains a 2.

Consider 100 through 399, which gives 300 numbers.

Count numbers in this range that do not contain 2:

- Hundreds digit: 1 or 3, giving 2 choices.
- Tens digit: any digit except 2, giving 9 choices.
- Units digit: any digit except 2, giving 9 choices.

So the number without a 2 is
\[
2\cdot9\cdot9=162
\]

Therefore the number containing at least one 2 is
\[
300-162=138
\]

ANSWER 7: C

Problem 8:  
Place the square with corners \((0,0)\), \((10,0)\), \((10,10)\), and \((0,10)\). Suppose the lemming starts at \((0,0)\) and runs along the diagonal toward \((10,10)\).

After running 6.2 m along the diagonal, its coordinates are
\[
\left(\frac{6.2}{\sqrt2},\frac{6.2}{\sqrt2}\right)
=
(3.1\sqrt2,3.1\sqrt2)
\]

A right turn sends it in a direction perpendicular to the diagonal. Running 2 more meters changes its coordinates by \((\sqrt2,-\sqrt2)\). So its final position is
\[
(4.1\sqrt2,2.1\sqrt2)
\]
which is still inside the square.

For any point inside a 10-by-10 square, the distances to the left and right sides add to 10, and the distances to the bottom and top sides add to 10. Thus the sum of the four distances is
\[
10+10=20
\]

The average of the four distances is
\[
\frac{20}{4}=5
\]

ANSWER 8: C

Problem 9:  
Reading the array as a \(5\times5\) grid:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

Find the smallest number in each row:

- Row 1: 2
- Row 2: 7
- Row 3: 3
- Row 4: 1
- Row 5: 2

Find the largest number in each column:

- Column 1: 13
- Column 2: 7
- Column 3: 15
- Column 4: 12
- Column 5: 9

The only number that is both a row minimum and a column maximum is 7, located in row 2, column 2.

ANSWER 9: C

Problem 10:  
Let Laila’s score on each of the first four tests be \(x\), and let her score on the last test be \(y\).

Her average was 82, so her total score was
\[
5\cdot82=410
\]

Thus
\[
4x+y=410
\]
so
\[
y=410-4x
\]

The scores are integers from 0 to 100 inclusive, and the last test score is higher than the first four, so we need:

1. \(y\le100\):
\[
410-4x\le100
\]
\[
310\le4x
\]
\[
x\ge78
\]

2. \(y\ge0\):
\[
410-4x\ge0
\]
\[
x\le102.5
\]
Since \(x\le100\), this is automatically satisfied.

3. \(y>x\):
\[
410-4x>x
\]
\[
410>5x
\]
\[
x<82
\]

So
\[
78\le x<82
\]
with \(x\) an integer, giving
\[
x=78,79,80,81
\]

These give last-test scores:
\[
y=98,94,90,86
\]
All are valid.

There are 4 possible values for the last test score.

ANSWER 10: A

# Problem 1:
Total possible 4-digit passwords (digits 0–9, repeats allowed): 10 × 10 × 10 × 10 = 10,000.

Passwords that **begin** with 9, 1, 1: the first three digits are forced, the fourth is free → 10 such passwords.

Allowed passwords: 10,000 − 10 = 9,990.

Ruling out others: 9000 would come from wrongly banning 1000 passwords; 9999 comes from removing only one password; 7290 comes from wrongly forbidding digits entirely.

ANSWER 1: D

# Problem 2:
Total weight of boys = 6 × 150 = 900 pounds.
Total weight of girls = 4 × 120 = 480 pounds.
Total = 1380 pounds for 10 children.

Average = 1380 ÷ 10 = 138 pounds.

(Note: 135 is the *unweighted* average of 150 and 120 — a trap, since there are more boys than girls, so the average must be above 135.)

ANSWER 2: C

# Problem 3:
Doubling each of the three dimensions multiplies the volume by 2 × 2 × 2 = 8.

Jellybeans ≈ 125 × 8 = 1000.

(250 would be just doubling once; 500 quadruples — both ignore that all three dimensions grow.)

ANSWER 3: E

# Problem 4:
Divide: 4 ÷ 37 = 0.108108108… so the repeating block is "108" of length 3.

Digit positions: position 1 → 1, position 2 → 0, position 3 → 8, then repeat.

100 ÷ 3 = 33 remainder 1, so the 100th digit is the **1st** digit of the block: 1.

ANSWER 4: B

# Problem 5:
Ratio 2 : 3 means 5 equal parts; 30 ÷ 5 = 6 students per part.

Boys = 2 × 6 = 12, Girls = 3 × 6 = 18.

Difference = 18 − 12 = 6. (Equivalently, the difference is 1 part = 6.)

ANSWER 5: D

# Problem 6:
Convert to minutes: 1 hr 15 min = 75 min; 1 hr 30 min = 90 min.

Minutes so far: 5(75) + 3(90) = 375 + 270 = 645.

To average 85 minutes over 9 days, total needed = 9 × 85 = 765.

Ninth day = 765 − 645 = 120 minutes = 2 hours.

ANSWER 6: E

# Problem 7:
Consider the numbers 101 through 399 (100 and 400 contain no digit 2, so counting 100–399 is equivalent).

- 100–199: last two digits must avoid 2 to *not* contain a 2 → 9 × 9 = 81 without a 2, so 100 − 81 = 19 contain a 2.
- 200–299: all 100 numbers contain a 2 (hundreds digit).
- 300–399: same as the 100s → 19 numbers.

Total: 19 + 100 + 19 = 138.

ANSWER 7: C

# Problem 8:
Key fact: for **any** point inside a square of side 10, the distances to one pair of opposite sides add to 10, and the distances to the other pair also add to 10.

So the four distances always total 10 + 10 = 20, no matter where the lemming ends up.

Average = 20 ÷ 4 = 5.

(The 6.2 m and 2 m are irrelevant distractors — hence 6.2 and 2 are traps.)

ANSWER 8: C

# Problem 9:
Write the array:

| | C1 | C2 | C3 | C4 | C5 |
|---|---|---|---|---|---|
|R1| 10 | 6 | 4 | 3 | 2 |
|R2| 11 | 7 | 14 | 10 | 8 |
|R3| 8 | 3 | 4 | 5 | 9 |
|R4| 13 | 4 | 15 | 12 | 1 |
|R5| 8 | 2 | 5 | 9 | 3 |

Column maxima: C1 = 13, C2 = 7, C3 = 15, C4 = 12, C5 = 9.

Now check whether each is the smallest in its row:
- 13 (R4): row 4 has 1 → no.
- 7 (R2): row 2 is 11, 7, 14, 10, 8 → 7 is smallest ✓
- 15 (R4): row min is 1 → no.
- 12 (R4): row min is 1 → no.
- 9 (R3): row 3 has 3 → no.

Also 6 and 1 aren't column maxima (C2's max is 7; C5's max is 9).

ANSWER 9: C

# Problem 10:
Let x be the score on each of the first four tests and y the last score.

4x + y = 5 × 82 = 410, with y > x and 0 ≤ x, y ≤ 100.

From 4x = 410 − y, we need 410 − y divisible by 4, so y ≡ 410 ≡ 2 (mod 4).

Condition y > x: y > (410 − y)/4 → 4y > 410 − y → 5y > 410 → y > 82.

So y is between 83 and 100 with y ≡ 2 (mod 4): y = 86, 90, 94, 98.

Check: y = 86 → x = 81 ✓; y = 98 → x = 78 ✓. That's 4 values.

ANSWER 10: A

Problem 1:

We are asked to count four-digit ATM passwords using digits \(0\) through \(9\), where repetition is allowed, but the password cannot begin with \(911\).

There are 10 choices for each of the 4 digits, so without restrictions there are

\[
10^4=10000
\]

possible passwords.

The forbidden passwords begin with \(9,1,1\). Their fourth digit can be anything from 0 through 9, so there are

\[
10
\]

forbidden passwords.

Thus the number allowed is

\[
10000-10=9990.
\]

Choice E, \(9999\), would subtract only one forbidden password, but there are 10 possible passwords beginning with \(911\). Choice C, \(9000\), incorrectly treats all passwords beginning with 9 as forbidden.

\[
\boxed{9990}
\]

ANSWER 1: D

Problem 2:

We are asked for the average weight of all 10 children.

The 6 boys have average weight 150 pounds, so their total weight is

\[
6\cdot150=900
\]

pounds.

The 4 girls have average weight 120 pounds, so their total weight is

\[
4\cdot120=480
\]

pounds.

Therefore, the total weight of all 10 children is

\[
900+480=1380
\]

pounds.

Their average is

\[
\frac{1380}{10}=138.
\]

The answer is not \(135\), which would be the simple average of 150 and 120; there are more boys than girls, so the combined average should be closer to 150.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 2: C

Problem 3:

Carrie’s box is twice as high, twice as wide, and twice as long as Bert’s box.

Volume depends on all three dimensions. Therefore the volume is multiplied by

\[
2\cdot2\cdot2=8.
\]

Bert’s box holds 125 jellybeans, so Carrie’s box holds approximately

\[
125\cdot 8=1000
\]

jellybeans.

Choices such as 250 or 500 account for only one or two dimensions being doubled. Since all three dimensions double, the capacity is eight times as great.

\[
\boxed{1000}
\]

ANSWER 3: E

Problem 4:

We need the 100th digit after the decimal point in

\[
\frac{4}{37}.
\]

Since

\[
\frac{1}{37}=0.\overline{027},
\]

we have

\[
\frac{4}{37}=0.\overline{108}.
\]

Thus the decimal digits repeat in the pattern

\[
1,0,8,1,0,8,\ldots
\]

The repeating block has length 3. Divide 100 by 3:

\[
100=3\cdot 33+1.
\]

So the 100th digit is the first digit of the repeating block \(108\), which is \(1\).

A tempting error is to use the third digit because 100 is large, but the remainder when dividing by 3 determines the position in the cycle.

\[
\boxed{1}
\]

ANSWER 4: B

Problem 5:

The ratio of boys to girls is

\[
2:3.
\]

There are

\[
2+3=5
\]

total ratio parts. Since there are 30 students, each part represents

\[
\frac{30}{5}=6
\]

students.

So the number of boys is

\[
2\cdot6=12,
\]

and the number of girls is

\[
3\cdot6=18.
\]

The number of additional girls is

\[
18-12=6.
\]

Choice A, 10, is the number of students in neither group calculation; the ratio difference is only one part, and each part is 6 students.

\[
\boxed{6}
\]

ANSWER 5: D

Problem 6:

We are asked how long Gage must skate on the ninth day to average 85 minutes per day over 9 days.

For the first 5 days, he skates 1 hour 15 minutes, or

\[
75\text{ minutes}.
\]

Thus, in those 5 days he skates

\[
5\cdot75=375
\]

minutes.

For the next 3 days, he skates 1 hour 30 minutes, or

\[
90\text{ minutes}.
\]

Thus, in those 3 days he skates

\[
3\cdot90=270
\]

minutes.

After 8 days, his total skating time is

\[
375+270=645
\]

minutes.

To average 85 minutes for 9 days, he needs a total of

\[
9\cdot85=765
\]

minutes.

Therefore, on the ninth day he must skate

\[
765-645=120
\]

minutes, which is

\[
2\text{ hours}.
\]

Choice C, 1 hour 20 minutes, would be only 80 minutes and would not raise the average to 85 minutes.

\[
\boxed{2\text{ hr}}
\]

ANSWER 6: E

Problem 7:

We need to count whole numbers from 100 through 399 that contain the digit 2.

There are

\[
399-100+1=300
\]

numbers from 100 through 399.

It is easier to count the numbers that do **not** contain a 2.

- The hundreds digit can be 1 or 3: 2 choices.
- The tens digit can be any digit except 2: 9 choices.
- The ones digit can be any digit except 2: 9 choices.

So the number containing no digit 2 is

\[
2\cdot9\cdot9=162.
\]

Therefore, the number containing at least one digit 2 is

\[
300-162=138.
\]

A common mistake is to count numbers with a 2 in each position separately, but that double-counts numbers such as 222 or 123.

\[
\boxed{138}
\]

ANSWER 7: C

Problem 8:

We are asked for the average of the lemming’s distances from the four sides of a square with side length 10 meters.

For any point inside a 10-by-10 square:

- Its distances from the left and right sides add to 10.
- Its distances from the bottom and top sides also add to 10.

Therefore, the sum of its distances from all four sides is always

\[
10+10=20.
\]

So the average of the four distances is

\[
\frac{20}{4}=5.
\]

The lemming’s path—running 6.2 meters along a diagonal and then turning—is extra information. Its final location changes, but the average of the distances to all four sides remains 5 as long as it is inside the square.

Choice D, 6.2, is the distance run along the diagonal, not the requested average distance to the sides.

\[
\boxed{5}
\]

ANSWER 8: C

Problem 9:

Arrange the numbers in the given \(5\times5\) array:

\[
\begin{array}{ccccc}
10&6&4&3&2\\
11&7&14&10&8\\
8&3&4&5&9\\
13&4&15&12&1\\
8&2&5&9&3
\end{array}
\]

We seek a number that is both:

1. the smallest number in its row, and  
2. the largest number in its column.

First find the smallest entry in each row:

- Row 1: smallest is \(2\)
- Row 2: smallest is \(7\)
- Row 3: smallest is \(3\)
- Row 4: smallest is \(1\)
- Row 5: smallest is \(2\)

Now test these candidates in their columns.

The \(7\) is in column 2. Column 2 is

\[
6,\ 7,\ 3,\ 4,\ 2.
\]

The largest number in this column is \(7\). Also, as found above, \(7\) is the smallest number in its row.

Thus \(7\) meets both requirements.

For example, \(15\) is large in its column, but it is not the smallest in its row. The number \(1\) is smallest in its ro