🔬 Model Lab

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre

Problem 1:
We are asked to find the number of integers between 1000 and 9999 that have four distinct digits.

To solve this, let's break down the process of forming a 4-digit number with distinct digits.

1. The first digit cannot be zero, so there are 9 choices (1 through 9).
2. The second digit can be any digit except the one chosen for the first digit, so there are 9 choices (0 and the 8 remaining digits).
3. The third digit can be any digit except the two chosen for the first and second digits, so there are 8 choices.
4. The fourth digit can be any digit except the three chosen for the first, second, and third digits, so there are 7 choices.

Now, we multiply the number of choices for each digit: $9 \times 9 \times 8 \times 7 = 4536$.

Let's rule out the wrong choices:
- A. 3024 is too low.
- C. 5040 is the number of permutations of 4 digits out of 10 without the restriction that the first digit cannot be zero.
- D. 6480 is too high.
- E. 6561 is $3^8$, which is unrelated to our calculation.

Thus, the correct answer is B. 4536.

ANSWER 1: B

Problem 2:
Maria works 8 hours a day, not including a 45-minute lunch break. She starts work at 7:25 A.M. and takes her lunch at noon. We need to find out when she ends her working day.

1. First, let's calculate the time she works before lunch:
   - From 7:25 A.M. to noon is 4 hours and 35 minutes.

2. Since she takes a 45-minute lunch break at noon, she resumes work at 12:45 P.M.

3. She needs to work a total of 8 hours. So, after lunch, she still needs to work $8 - 4$ hours and $35$ minutes, which is $3$ hours and $25$ minutes.

4. Adding this to her resume time:
   - From 12:45 P.M. to 4:10 P.M. is 3 hours and 25 minutes.

Thus, her working day ends at 4:10 P.M.

Let's rule out the wrong choices:
- A. 3:40 P.M. is too early.
- B. 3:55 P.M. is too early.
- D. 4:25 P.M. is too late.
- E. 4:40 P.M. is too late.

Thus, the correct answer is C. 4:10 P.M.

ANSWER 2: C

Problem 3:
Harold ate $\frac{1}{4}$ of the pie, leaving $\frac{3}{4}$ of it. Then, a moose ate $\frac{1}{3}$ of what Harold left, which is $\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ of the original pie. So, after the moose ate, $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ of the original pie remained.

After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind, which is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the original pie. So, after the porcupine ate, $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ of the original pie remained.

Let's rule out the wrong choices:
- A. $\frac{1}{12}$ is too low.
- B. $\frac{1}{6}$ is too low.
- C. $\frac{1}{4}$ is too low.
- E. $\frac{5}{12}$ is too low.

Thus, the correct answer is D. $\frac{1}{3}$.

ANSWER 3: D

Problem 4:
The arithmetic mean of four numbers is 85. So, the sum of these four numbers is $85 \times 4 = 340$. If the largest number is 97, then the sum of the remaining three numbers is $340 - 97 = 243$.

The mean of these three numbers is $\frac{243}{3} = 81$.

Let's rule out the wrong choices:
- B. 82.7 is too high.
- C. 83.0 is too high.
- D. 84.0 is too high.
- E. 84.3 is too high.

Thus, the correct answer is A. 81.0.

ANSWER 4: A

Problem 5:
Let's denote the side length of the smaller square as $s$ and the larger square as $S$. The perimeter of the smaller square is $4s$, and the perimeter of the larger square is $4S$. We are given that $4S = 3 \times 4s$, so $S = 3s$.

The area of the smaller square is $s^2$, and the area of the larger square is $S^2 = (3s)^2 = 9s^2$.

Thus, the area of the larger square is 9 times the area of the smaller square.

Let's rule out the wrong choices:
- A. 2 is too low.
- B. 3 is too low.
- C. 4 is too low.
- D. 6 is too low.

Thus, the correct answer is E. 9.

ANSWER 5: E

Problem 6:
To find the ratio of the least common multiple (LCM) of 180 and 594 to their greatest common factor (GCF), we first need to calculate the GCF and LCM.

1. Prime factorization of 180: $2^2 \times 3^2 \times 5$
2. Prime factorization of 594: $2 \times 3^3 \times 11$

GCF is found by taking the lowest power of common primes: $2 \times 3^2 = 18$.

LCM is found by taking the highest power of all primes involved: $2^2 \times 3^3 \times 5 \times 11 = 5940$.

The ratio is $\frac{LCM}{GCF} = \frac{5940}{18} = 330$.

Let's rule out the wrong choices:
- A. 110 is too low.
- B. 165 is too low.
- D. 625 is too high and not related to our calculation.
- E. 660 is too high.

Thus, the correct answer is C. 330.

ANSWER 6: C

Problem 7:
Given $a ♦ b = \frac{a + b}{a - b}$, we need to find $(6 ♦ 4) ♦ 3$.

1. First, calculate $6 ♦ 4 = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5$.
2. Then, calculate $5 ♦ 3 = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.

Let's rule out the wrong choices:
- B. 13 is too high.
- C. 15 is too high.
- D. 30 is too high.
- E. 72 is too high.

Thus, the correct answer is A. 4.

ANSWER 7: A

Problem 8:
Tori's test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry. She answered 70% of arithmetic, 40% of algebra, and 60% of geometry correctly.

1. Correct arithmetic answers: $0.7 \times 10 = 7$.
2. Correct algebra answers: $0.4 \times 30 = 12$.
3. Correct geometry answers: $0.6 \times 35 = 21$.

Total correct answers: $7 + 12 + 21 = 40$.

To pass, she needs $60\%$ of 75, which is $0.6 \times 75 = 45$ correct answers. So, she needs $45 - 40 = 5$ more correct answers.

Let's rule out the wrong choices:
- A. 1 is too low.
- C. 7 is too high.
- D. 9 is too high.
- E. 11 is too high.

Thus, the correct answer is B. 5.

ANSWER 8: B

Problem 9:
Six trees are equally spaced along a straight road. The distance from the first tree to the fourth tree is 60 feet. This distance covers 3 intervals between trees.

So, the distance between two consecutive trees is $\frac{60}{3} = 20$ feet.

The distance between the first and last trees covers 5 intervals, so it is $5 \times 20 = 100$ feet.

Let's rule out the wrong choices:
- A. 90 is too low.
- C. 105 is too high.
- D. 120 is too high.
- E. 140 is too high.

Thu

Problem 1:

We need the number of 4-digit integers from 1000 to 9999 with four distinct digits.

A 4-digit number cannot start with 0.

- Thousands digit: 9 choices, digits 1 through 9.
- Hundreds digit: 9 choices, any digit except the thousands digit, including possibly 0.
- Tens digit: 8 choices remaining.
- Ones digit: 7 choices remaining.

So the total is

\[
9 \times 9 \times 8 \times 7 = 4536.
\]

Choice C, 5040, would be \(10 \times 9 \times 8 \times 7\), but that incorrectly allows the first digit to be 0.

ANSWER 1: B

Problem 2:

Maria must work 8 hours, not including her 45-minute lunch.

She starts at 7:25 A.M. and works until noon.

From 7:25 A.M. to noon is:

\[
4\text{ hours }35\text{ minutes}
\]

She needs 8 hours total, so after lunch she still must work:

\[
8:00 - 4:35 = 3:25
\]

Her lunch is 45 minutes, so from noon to 12:45 P.M. she is at lunch.

Then she works 3 hours 25 minutes after 12:45 P.M.:

\[
12:45 + 3:25 = 4:10\text{ P.M.}
\]

ANSWER 2: C

Problem 3:

Harold ate \(\frac14\) of the pie, so he left:

\[
1 - \frac14 = \frac34
\]

The moose ate \(\frac13\) of what Harold left:

\[
\frac13 \cdot \frac34 = \frac14
\]

So after the moose, the amount remaining was:

\[
\frac34 - \frac14 = \frac12
\]

The porcupine ate \(\frac13\) of what the moose left:

\[
\frac13 \cdot \frac12 = \frac16
\]

So the amount still remaining was:

\[
\frac12 - \frac16 = \frac36 - \frac16 = \frac26 = \frac13
\]

ANSWER 3: D

Problem 4:

The average of four numbers is 85, so their total sum is:

\[
4 \times 85 = 340
\]

The largest number is 97. Remove it from the total:

\[
340 - 97 = 243
\]

Now find the average of the remaining three numbers:

\[
\frac{243}{3} = 81
\]

ANSWER 4: A

Problem 5:

The perimeter of a square is \(4s\), where \(s\) is the side length.

If the perimeter of the larger square is 3 times the perimeter of the smaller square, then the side length of the larger square is also 3 times the side length of the smaller square.

Area depends on the square of the side length:

\[
3^2 = 9
\]

So the larger square’s area is 9 times the smaller square’s area.

ANSWER 5: E

Problem 6:

We need the ratio:

\[
\frac{\operatorname{lcm}(180,594)}{\gcd(180,594)}
\]

Prime factorize:

\[
180 = 2^2 \cdot 3^2 \cdot 5
\]

\[
594 = 2 \cdot 3^3 \cdot 11
\]

The greatest common factor uses the smaller powers:

\[
\gcd(180,594)=2^1 \cdot 3^2 = 18
\]

The least common multiple uses the larger powers:

\[
\operatorname{lcm}(180,594)=2^2 \cdot 3^3 \cdot 5 \cdot 11
\]

\[
=4 \cdot 27 \cdot 55 = 5940
\]

Now compute the ratio:

\[
\frac{5940}{18}=330
\]

ANSWER 6: C

Problem 7:

The operation is defined by:

\[
a \diamond b = \frac{a+b}{a-b}
\]

First find \(6 \diamond 4\):

\[
6 \diamond 4 = \frac{6+4}{6-4} = \frac{10}{2}=5
\]

Now compute \((6 \diamond 4) \diamond 3 = 5 \diamond 3\):

\[
5 \diamond 3 = \frac{5+3}{5-3} = \frac{8}{2}=4
\]

ANSWER 7: A

Problem 8:

Tori’s test had 75 problems.

She got:

Arithmetic:

\[
70\% \text{ of } 10 = 0.70 \times 10 = 7
\]

Algebra:

\[
40\% \text{ of } 30 = 0.40 \times 30 = 12
\]

Geometry:

\[
60\% \text{ of } 35 = 0.60 \times 35 = 21
\]

Total correct:

\[
7+12+21=40
\]

To pass with 60%, she needed:

\[
60\% \text{ of } 75 = 0.60 \times 75 = 45
\]

So she needed:

\[
45-40=5
\]

more correct answers.

ANSWER 8: B

Problem 9:

Six trees are equally spaced.

From the first tree to the fourth tree, there are 3 equal spaces:

- first to second
- second to third
- third to fourth

These 3 spaces total 60 feet, so one space is:

\[
\frac{60}{3}=20
\]

From the first tree to the last, meaning the sixth tree, there are 5 equal spaces.

So the distance is:

\[
5 \times 20 = 100
\]

ANSWER 9: B

Problem 10:

As written, this problem is missing the reading rates or times for Alice, Bob, and Chandra. In the standard version of this problem, Alice, Bob, and Chandra can read the whole book in 4 hours, 6 hours, and 12 hours respectively.

Their combined reading rate is:

\[
\frac14+\frac16+\frac1{12}
\]

Using denominator 12:

\[
\frac3{12}+\frac2{12}+\frac1{12}=\frac6{12}=\frac12
\]

So together, they read \(\frac12\) of the book per hour. Therefore, they finish the book in:

\[
2 \text{ hours}
\]

Convert 2 hours to seconds:

\[
2 \times 60 \times 60 = 7200
\]

ANSWER 10: E

Problem 11:

We need simplify:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

Notice that:

\[
19.98 = 10 \times 1.998
\]

So:

\[
100 \times 19.98 \times 1.998 \times 1000
\]

\[
=100 \times (10 \times 1.998) \times 1.998 \times 1000
\]

\[
=100 \times 10 \times 1000 \times (1.998)^2
\]

\[
=1{,}000{,}000 \times (1.998)^2
\]

Since:

\[
1{,}000{,}000 = 1000^2
\]

we have:

\[
1{,}000{,}000 \times (1.998)^2 = (1000 \times 1.998)^2
\]

\[
= (1998)^2
\]

ANSWER 11: D

Problem 12:

Use the order of operations.

First do the parentheses:

\[
6 \div 3 = 2
\]

Then multiply:

\[
9 \times 2 = 18
\]

Now add:

\[
-15 + 18 = 3
\]

ANSWER 12: D

Problem 1:
We are asked to find the number of integers between 1000 and 9999 (inclusive) that have four distinct digits. 
A four-digit integer has four place values: thousands, hundreds, tens, and units. We can determine the number of choices for each digit step-by-step:
1. **Thousands digit:** It can be any digit from 1 to 9 (it cannot be 0, otherwise it wouldn't be a four-digit number). This gives us **9 choices**.
2. **Hundreds digit:** It can be any digit from 0 to 9, except the one already used for the thousands digit. This gives us 10 - 1 = **9 choices**.
3. **Tens digit:** It can be any digit from 0 to 9, except the two already used. This gives us 10 - 2 = **8 choices**.
4. **Units digit:** It can be any digit from 0 to 9, except the three already used. This gives us 10 - 3 = **7 choices**.

To find the total number of such integers, we multiply the number of choices for each digit:
Total = 9 × 9 × 8 × 7
Total = 81 × 56
Total = 4536

Let's rule out the other choices:
- A (3024) is 9 × 8 × 7 × 6, which incorrectly assumes the hundreds digit only has 8 choices.
- C (5040) is 10 × 9 × 8 × 7, which incorrectly allows the thousands digit to be 0.
- D (6480) is 9 × 9 × 8 × 10, which doesn't restrict the units digit.
- E (6561) is 9 × 9 × 9 × 9, which allows repeated digits.

ANSWER 1: B

Problem 2:
We are asked to find the time Maria's working day ends, given her start time, work duration, and lunch break.
Maria's total time at work includes her 8 hours of actual work plus her 45-minute lunch break. 
Total time at work = 8 hours and 45 minutes.

She begins working at 7:25 A.M. We can add her total time at work in two steps:
1. Add the 8 hours to 7:25 A.M.:
7:25 A.M. + 8 hours = 3:25 P.M. (since 7 + 8 = 15, and 15:25 in 12-hour time is 3:25 P.M.)
2. Add the 45 minutes to 3:25 P.M.:
25 minutes + 45 minutes = 70 minutes.
Since 60 minutes is 1 hour, 70 minutes is 1 hour and 10 minutes.
So, 3:00 P.M. + 1 hour and 10 minutes = 4:10 P.M.

Let's rule out the other choices:
- A (3:40 P.M.) forgets to add the 45-minute lunch break and incorrectly adds 15 minutes instead.
- B (3:55 P.M.) incorrectly adds only a 30-minute lunch break.
- D (4:25 P.M.) incorrectly adds a full 1-hour lunch break instead of 45 minutes.
- E (4:40 P.M.) incorrectly adds the 45 minutes to 3:55 P.M.

ANSWER 2: C

Problem 3:
We are asked to find the fraction of the original pie that remains after Harold, a moose, and a porcupine each eat a portion of what is left.
Let the original pie be 1 whole.
1. **Harold** eats 1/4 of the pie. 
   The fraction remaining is 1 - 1/4 = 3/4.
2. **The moose** eats 1/3 of what Harold left behind. This means the moose leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (3/4) = 6/12 = 1/2.
3. **The porcupine** eats 1/3 of what the moose left behind. This means the porcupine leaves behind 2/3 of what it found.
   The fraction remaining is (2/3) × (1/2) = 2/6 = 1/3.

Let's rule out the other choices:
- A (1/12) is the result of multiplying 1/4 × 1/3 × 1/3, which is the fraction the porcupine ate, not what remained.
- B (1/6) is the result of 1/4 × 2/3.
- C (1/4) is the amount Harold ate.
- E (5/12) is the result of subtracting 1/4 and 1/3 directly from 1 whole (1 - 1/4 - 1/3 = 5/12), which ignores that the animals ate fractions of the *remainder*, not the whole pie.

ANSWER 3: D

Problem 4:
We are asked to find the mean of three numbers, given the mean of four numbers and the value of the largest number.
The arithmetic mean of four numbers is 85. 
Total sum of the four numbers = 4 × 85 = 340.

The largest of these numbers is 97. We subtract this from the total sum to find the sum of the remaining three numbers:
Sum of the remaining three numbers = 340 - 97 = 243.

To find the mean of these remaining three numbers, we divide their sum by 3:
Mean = 243 / 3 = 81.0.

Let's rule out the other choices:
- B (82.7) is roughly (340/4) - (97/4).
- C (83.0) and D (84.0) are just arbitrary subtractions from the original mean.
- E (84.3) comes from an arithmetic error: doing 340 - 97 = 253, and then 253 / 3 ≈ 84.3.

ANSWER 4: A

Problem 5:
We are asked to find the ratio of the area of a larger square to the area of a smaller square, given that the perimeter of the larger square is 3 times that of the smaller square.
Let the side length of the smaller square be $s$. 
Its perimeter is $4s$, and its area is $s^2$.

The perimeter of the larger square is 3 times the perimeter of the smaller square:
Perimeter of larger square = $3 \times 4s = 12s$.
The side length of the larger square is $12s / 4 = 3s$.
The area of the larger square is $(3s)^2 = 9s^2$.

The ratio of the area of the larger square to the area of the smaller square is:
$9s^2 / s^2 = 9$.
*(Rule of thumb: If the ratio of the perimeters (or any linear dimension) of two similar figures is $k$, the ratio of their areas is $k^2$. Here $k=3$, so $k^2=9$.)*

Let's rule out the other choices:
- A (2) is incorrect.
- B (3) is the ratio of their perimeters and side lengths, not their areas.
- C (4) would be the area ratio if the perimeter ratio was 2.
- D (6) incorrectly multiplies the ratio by 2 instead of squaring it.

ANSWER 5: E

Problem 6:
We are asked to find the ratio of the least common multiple (LCM) to the greatest common factor (GCF) of 180 and 594.
First, find the prime factorizations of both numbers:
180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5
594 = 2 × 297 = 2 × 3 × 99 = 2 × 3 × 9 × 11 = 2¹ × 3³ × 11

Next, find the GCF by taking the lowest power of each common prime factor:
GCF = 2¹ × 3² = 2 × 9 = 18

Then, find the LCM by taking the highest power of all prime factors present:
LCM = 2² × 3³ × 5 × 11 = 4 × 27 × 5 × 11 = 5940

Finally, find the ratio of the LCM to the GCF:
Ratio = LCM / GCF = 5940 / 18 = 330.
*(Alternatively, LCM / GCF = (2² × 3³ × 5 × 11) / (2¹ × 3²) = 2¹ × 3¹ × 5 × 11 = 330)*

Let's rule out the other choices:
- A (110) misses the factor of 3.
- B (165) misses the factor of 2.
- D (625) is completely unre