🔬 Model Lab

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len

Problem 1:
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Let's denote Bonnie's speed as $v$ meters per second. Then Annie's speed is $1.25v$ meters per second because she runs 25% faster. To catch up to Bonnie, Annie needs to be one full lap (400 meters) ahead of her. The relative speed between Annie and Bonnie is $1.25v - v = 0.25v$ meters per second. The time it takes for Annie to be 400 meters ahead is $400 / 0.25v = 1600 / v$ seconds. In this time, Annie covers a distance of $1.25v \times (1600 / v) = 2000$ meters. Since the track is 400 meters, the number of laps Annie runs is $2000 / 400 = 5$. Let's check the other options: if Annie runs 4 laps (1600 meters), the time taken is $1600 / (1.25v) = 1280 / v$, and Bonnie covers $v \times (1280 / v) = 1280$ meters, so Annie is $1600 - 1280 = 320$ meters ahead, not enough to pass Bonnie. Thus, the correct number of laps is indeed more than 4. Checking other options directly isn't necessary as we've derived the correct answer.
ANSWER 1: D

Problem 2:
Last week small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
First, let's find the price per box last week and this week. Last week, the price per box was $5/4 = $1.25. This week, the price per box is $4/5 = $0.80. The decrease in price is $1.25 - $0.80 = $0.45. To find the percent decrease, we use the formula: $(\text{decrease} / \text{original price}) \times 100 = ($0.45 / $1.25) \times 100 = 36%. The closest answer is 35%.
Let's evaluate the other choices: 30% would be a decrease to $0.875 (too high), and 40% would be a decrease to $0.75 (too low), so 35% is the closest.
ANSWER 2: B

Problem 3:
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?
First, let's count the numbers divisible by 3: $3, 6, ..., 45$. The sequence starts at 3, ends at 45, and increases by 3. The number of terms is $(45 - 3)/3 + 1 = 15$. Next, count the numbers divisible by 5: $5, 10, ..., 45$. The number of terms is $(45 - 5)/5 + 1 = 9$. However, numbers divisible by both 3 and 5 (i.e., 15) are counted twice. The numbers divisible by 15 are $15, 30, 45$, so there are 3 such numbers. Using the principle of inclusion-exclusion, the total count is $15 + 9 - 3 = 21$.
Let's check the other options: counting directly or using other methods isn't necessary as we've derived the correct count.
ANSWER 3: B

Problem 4:
9 7 × 53 =
To solve this, we need to understand that $9.7 \times 5.3 = 51.41$. Let's examine the choices by converting them into the same form: 
A. $.9 .7 × 53 = 0.97 \times 53 = 51.41$
B. $.9 .7 × .53 = 0.97 \times 0.53 = 0.5141$
C. $.9 .7 × 5.3 = 0.97 \times 5.3 = 5.141$
D. $.9 7 × .53 = 0.97 \times 0.53 = 0.5141$
E. $.09 .07 × .53 = 0.0907 \times 0.53 = 0.048071$
Only A gives the same result as $9.7 \times 5.3$.
ANSWER 4: A

Problem 5:
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Let's denote the number of tiles on one side of the square floor as $n$. The number of tiles on the two diagonals is $2n - 1$ because the center tile is counted twice when adding the diagonals. We're given that $2n - 1 = 37$, so $2n = 38$, and $n = 19$. The total number of tiles is $n^2 = 19^2 = 361$.
Let's verify: if $n = 18$, then $2n - 1 = 35$ (too low), and if $n = 20$, then $2n - 1 = 39$ (too high), so $n = 19$ is correct.
ANSWER 5: C

Problem 6:
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is 2 miles, which is 10,560 feet, and Bella covers 2½ feet with each step. How many steps will Bella take by the time she meets Ella?
Let's denote Bella's speed as $v$ feet per second. Then Ella's speed is $5v$ feet per second. Their combined speed is $v + 5v = 6v$ feet per second. The time it takes for them to meet is $10560 / (6v) = 1760 / v$ seconds. Bella's speed is $v$ feet per second, so in $1760 / v$ seconds, she covers $v \times (1760 / v) = 1760$ feet. Since Bella covers 2.5 feet per step, the number of steps she takes is $1760 / 2.5 = 704$.
Let's check: if Bella walks at a certain speed, Ella's speed is 5 times that. The ratio of their speeds is 1:5, so the ratio of the distances they cover is also 1:5. Bella covers $1/6$ of the total distance, which is $10560 / 6 = 1760$ feet, confirming our calculation.
ANSWER 6: A

Problem 7:
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
Let's denote the original length and width as $L$ and $W$, respectively. The original area is $L \times W$. After the increase, the new length is $1.2L$ and the new width is $1.5W$, so the new area is $1.2L \times 1.5W = 1.8LW$. The percent increase in area is $((1.8LW - LW) / LW) \times 100 = (0.8LW / LW) \times 100 = 80%$.
Let's verify: the new area is 1.8 times the original, which is an increase of 80%.
ANSWER 7: D

Problem 8:
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
To solve this, we need to factorize 6545. Let's start by checking divisibility by small primes: 6545 is divisible by 5 because it ends in 5, so $6545 = 5 \times 1309$. Further, $1309 = 7 \times 11 \times 17$ (after checking divisibility or using factorization methods). So, $6545 = 5 \times 7 \times 11 \times 17$. We need to combine these factors into two two-digit numbers. One way to do this is $5 \times 7 = 35$ and $11 \times 17 = 187$ (too high since 187 is three digits). Another way is $5 \times 11 

Problem 1:

Annie runs 25% faster than Bonnie, so Annie’s speed is

\[
1.25=\frac54
\]

times Bonnie’s speed.

They start together, and Annie first passes Bonnie when Annie has run exactly 1 more lap than Bonnie.

Suppose Bonnie has run \(x\) laps. Then Annie has run

\[
\frac54 x
\]

laps. We need

\[
\frac54 x - x = 1
\]

\[
\frac14 x = 1
\]

\[
x=4
\]

So Bonnie has run 4 laps, and Annie has run

\[
\frac54 \cdot 4 = 5
\]

laps.

Tempting wrong choices like 4 laps count Bonnie’s laps, not Annie’s. Annie has run 5 laps.

ANSWER 1: D

Problem 2:

Last week the price was 4 boxes for \$5, so the price per box was

\[
\frac{5}{4}=1.25
\]

dollars per box.

This week the price is 5 boxes for \$4, so the price per box is

\[
\frac{4}{5}=0.80
\]

dollars per box.

The decrease in price per box is

\[
1.25-0.80=0.45
\]

The percent decrease is based on the original price:

\[
\frac{0.45}{1.25}=0.36=36\%
\]

The closest answer is \(35\%\).

A tempting wrong answer is \(45\%\), but 45 cents is the amount of decrease, not the percent decrease.

ANSWER 2: B

Problem 3:

We want whole numbers from 1 through 46 divisible by 3 or 5 or both.

Multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But numbers divisible by both 3 and 5 are multiples of 15, and they were counted twice.

Multiples of 15:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

So by inclusion-exclusion:

\[
15+9-3=21
\]

ANSWER 3: B

Problem 4:

The expression is

\[
\frac{9}{7}\times 53
\]

Choice A is

\[
\frac{0.9}{0.7}\times 53
\]

But

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

because multiplying numerator and denominator by 10 gives

\[
\frac{0.9}{0.7}=\frac{9}{7}
\]

So Choice A is exactly equal to the original expression.

The other choices change the \(53\) into \(5.3\), \(0.53\), or change the fraction incorrectly, so they are not equal to the original expression.

ANSWER 4: A

Problem 5:

Suppose the floor is an \(n \times n\) square of tiles.

Each diagonal contains \(n\) tiles.

If \(n\) is odd, the two diagonals share the center tile, so the total number of tiles on the two diagonals is

\[
2n-1
\]

We are told this equals 37:

\[
2n-1=37
\]

\[
2n=38
\]

\[
n=19
\]

So the floor has

\[
19 \times 19 = 361
\]

tiles.

If \(n\) were even, the two diagonals would have \(2n\) tiles total, which could not equal 37, an odd number.

ANSWER 5: C

Problem 6:

Ella rides 5 times as fast as Bella walks.

So their speed ratio is

\[
\text{Bella} : \text{Ella} = 1:5
\]

Together, they cover the full distance between the houses, which is 10,560 feet.

Since Bella covers 1 part and Ella covers 5 parts, the total is 6 parts. Bella covers

\[
\frac{1}{6}\cdot 10560 = 1760
\]

feet.

Bella takes one step every \(2\frac12 = 2.5\) feet, so the number of steps is

\[
\frac{1760}{2.5}=704
\]

ANSWER 6: A

Problem 7:

Let the original rectangle have length \(L\) and width \(W\). Its original area is

\[
LW
\]

The length is increased by 20%, so the new length is

\[
1.2L
\]

The width is increased by 50%, so the new width is

\[
1.5W
\]

The new area is

\[
(1.2L)(1.5W)=1.8LW
\]

So the new area is 180% of the original area, meaning the area increased by

\[
180\%-100\%=80\%
\]

A common mistake is to add \(20\%+50\%=70\%\), but area depends on multiplying both dimensions.

ANSWER 7: D

Problem 8:

We need to write 6545 as a product of two positive two-digit numbers.

Factor 6545:

\[
6545 \div 5 = 1309
\]

Now factor 1309:

\[
1309 = 7 \cdot 187 = 7 \cdot 11 \cdot 17
\]

So

\[
6545 = 5 \cdot 7 \cdot 11 \cdot 17
\]

Now look for two two-digit factors. Group them as

\[
(7\cdot 11)(5\cdot 17)
\]

\[
77 \cdot 85 = 6545
\]

Both 77 and 85 are two-digit numbers.

Their sum is

\[
77+85=162
\]

ANSWER 8: A

Problem 9:

Sri’s car clock gains time at a constant rate.

When 30 actual minutes pass, the car clock goes from 12:00 to 12:35, so it shows 35 minutes have passed.

Thus the car clock runs at the ratio

\[
\frac{35}{30}=\frac76
\]

of actual time.

Later, the car clock says 7:00. Since it started at 12:00, the car clock shows 7 hours have passed:

\[
7 \text{ hours} = 420 \text{ minutes}
\]

Let the actual elapsed time be \(t\). Then

\[
\frac76 t = 420
\]

\[
t = 420 \cdot \frac67 = 360
\]

So 360 actual minutes, or 6 hours, have passed.

Starting from 12:00 noon, the actual time is

\[
6:00
\]

ANSWER 9: B

Problem 10:

Let the cost of one pamphlet be \(p\) dollars.

Nine copies cost less than \$10:

\[
9p<10
\]

\[
p<\frac{10}{9}\approx 1.111\ldots
\]

Ten copies cost more than \$11:

\[
10p>11
\]

\[
p>1.10
\]

So the price must be greater than \$1.10 but less than about \$1.111.

Check the choices:

- \$1.10 gives exactly \$11.00 for 10 copies, not more than \$11.00.
- \$1.11 gives \(10 \cdot 1.11 = 11.10\), which is more than \$11.00.
- Also \(9 \cdot 1.11 = 9.99\), which is less than \$10.00.

So the price is \$1.11.

ANSWER 10: E

Problem 11:

Let the regular price of one tire be \(r\) dollars.

The sale says: buy three tires at regular price and get the fourth tire for \$3.

So Sam paid

\[
3r+3=240
\]

Subtract 3:

\[
3r=237
\]

Divide by 3:

\[
r=79
\]

So the regular price of one tire was \$79.

ANSWER 11: D

Problem 12:

First, walking:

45 minutes is

\[
\frac{45}{60}=\frac34
\]

of an hour.

At 4 mph, the walking distance is

\[
4 \cdot \frac34 = 3
\]

miles.

Next, running:

30 minutes is

\[
\frac{30}{60}=\frac12
\]

of an hour.

At 10 mph, the running distance is

\[
10 \cdot \frac12 = 5
\]

miles.

Total distance:

\[
3+5=8
\]

miles.

ANSWER 12: B

**Problem 1:**

**Restate what is being asked:** We need to determine how many laps Annie will have completed at the exact moment she overtakes (laps) Bonnie for the first time.

**Set up the math:** Let Bonnie's speed be $v$. Since Annie runs 25% faster than Bonnie, Annie's speed is $1.25v$. 
When Annie passes Bonnie for the first time, she must have run exactly 1 more lap than Bonnie. Let $d_A$ be the number of laps Annie runs and $d_B$ be the number of laps Bonnie runs. We know:
1) $d_A = 1.25 d_B$ (since distance is proportional to speed for the same amount of time)
2) $d_A - d_B = 1$

**Computation:** 
Substitute the first equation into the second:
$1.25 d_B - d_B = 1$
$0.25 d_B = 1$
$d_B = 4$ laps

Bonnie has run 4 laps. To find Annie's laps, we add 1:
$d_A = 4 + 1 = 5$ laps.

**Rule out wrong choices:** 
- A (1 1/4) is the ratio of their speeds, not the number of laps.
- B (3 1/3) is a random fraction.
- C (4) is the number of laps *Bonnie* has run, not Annie.
- E (25) is the percentage given in the problem.

ANSWER 1: D

***

**Problem 2:**

**Restate what is being asked:** We need to find the approximate percentage decrease in the price of a single box of facial tissue during the sale.

**Set up the math:** 
First, find the price per box before and during the sale.
- Old price per box: $\$5 \div 4 = \$1.25$
- New price per box: $\$4 \div 5 = \$0.80$

The formula for percent decrease is: $\frac{\text{Old Price} - \text{New Price}}{\text{Old Price}} \times 100\%$

**Computation:** 
Decrease in price = $\$1.25 - \$0.80 = \$0.45$
Percent decrease = $\frac{0.45}{1.25} \times 100\% = \frac{45}{125} \times 100\% = \frac{9}{25} \times 100\% = 9 \times 4\% = 36\%$

The value 36% is closest to 35%.

**Rule out wrong choices:** 
- A (30%) would mean a decrease of $\$0.375$.
- C (40%) would mean a decrease of $\$0.50$.
- D (45%) would mean a decrease of $\$0.5625$.
- E (65%) is the remaining percentage of the price ($100\% - 35\%$).

ANSWER 2: B

***

**Problem 3:**

**Restate what is being asked:** We need to find the total number of integers from 1 to 46 inclusive that are multiples of 3, multiples of 5, or multiples of both.

**Set up the math:** We use the Principle of Inclusion-Exclusion. We will count the multiples of 3, add the multiples of 5, and subtract the multiples of 15 (since they are counted twice).

**Computation:** 
- Multiples of 3: $\lfloor \frac{46}{3} \rfloor = 15$
- Multiples of 5: $\lfloor \frac{46}{5} \rfloor = 9$
- Multiples of 15: $\lfloor \frac{46}{15} \rfloor = 3$

Total = $15 + 9 - 3 = 21$.

**Rule out wrong choices:** 
- A (18) is $15 + 3$, which adds the intersection instead of subtracting.
- C (24) is $15 + 9$, which forgets to subtract the overlapping multiples of 15.
- D (25) and E (27) are miscounts.

ANSWER 3: B

***

**Problem 4:**

**Restate what is being asked:** The text "9 7 × 53 =" is a typographical representation of the fraction multiplication $\frac{9}{7} \times 53$. We need to identify which of the multiple-choice options is mathematically equivalent to this expression.

**Set up the math:** We are looking for an expression that simplifies to exactly $\frac{9}{7} \times 53$. 

**Computation:** 
Let's evaluate Choice A: $\frac{0.9}{0.7} \times 53$. 
If we multiply the numerator and the denominator of the fraction by 10, we get:
$\frac{0.9 \times 10}{0.7 \times 10} \times 53 = \frac{9}{7} \times 53$.
This matches our target expression perfectly.

**Rule out wrong choices:** 
- B ($\frac{0.9}{0.7} \times 0.53$) multiplies by $0.53$ instead of $53$.
- C ($\frac{0.9}{0.7} \times 5.3$) multiplies by $5.3$ instead of $53$.
- D ($\frac{0.9}{7} \times 0.53$) changes the denominator to $7$ and multiplies by $0.53$.
- E ($\frac{0.09}{0.07} \times 0.53$) simplifies to $\frac{9}{7} \times 0.53$.

ANSWER 4: A

***

**Problem 5:**

**Restate what is being asked:** We are given a square floor made of a grid of smaller square tiles. The two main diagonals contain a total of 37 tiles. We need to find the total number of tiles on the floor.

**Set up the math:** Let the floor be an $n \times n$ grid. 
Each diagonal has exactly $n$ tiles. 
If $n$ is even, the diagonals do not share any tiles, so the total number of tiles on the diagonals would be $2n$. (This would mean $2n = 37$, which is impossible since $n$ must be a whole number).
If $n$ is odd, the two diagonals intersect and share exactly 1 center tile. The total number of tiles on the diagonals is $n + n - 1 = 2n - 1$.

**Computation:** 
$2n - 1 = 37$
$2n = 38$
$n = 19$

The floor is a $19 \times 19$ grid. 
Total tiles = $19 \times 19 = 361$.

**Rule out wrong choices:** 
- A (148) is simply $37 \times 4$.
- B (324) is $18^2$, which would have 36 diagonal tiles.
- D (1296) is $36^2$.
- E (1369) is $37^2$, assuming the side length was 37 instead of the diagonals.

ANSWER 5: C

***

**Problem 6:**

**Restate what is being asked:** Bella and Ella travel toward each other from 10,560 feet apart. Ella travels 5 times as fast as Bella. We need to find how many 2.5-foot steps Bella takes before they meet.

**Set up the math:** Because they travel for the same amount of time, the ratio of the distances they cover is equal to the ratio of their speeds. 
Let $d_B$ be Bella's distance. Ella's distance is $5d_B$.
The total distance is $d_B + 5d_B = 10,560$ feet.

**Computation:** 
$6d_B = 10,560$
$d_B = 1,760$ feet.

Bella walks 1,760 feet. Since each step is 2.5 feet (or $\frac{5}{2}$ feet), the number of steps is:
$1,760 \div 2.5 = 1,760 \times \frac{2}{5} = \frac{3,520}{5} = 704$ steps.

**Rule out wrong choices:** 
- B (845) and C (1056) are miscalculations.
- D (1760) is the distance Bella walks in feet, not steps.
- E (3520) is $1760 \times 2$, forgetting to divide by 5.

ANSWER 6: A

***

**Problem 7:**

**Restate what is being asked:** We need to find the percentage increase in the area of a rectangle when its length is increased by 20% and its width is increased by 50%.

**Set up the math:** Let the original len