🔬 Model Lab

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number

Problem 1:
Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Let's denote Brynn's original savings as $x$. 
In July, her savings decreased by 20%, so her savings became $x - 0.2x = 0.8x$.
In August, her savings increased by 50% of the new amount, so her savings became $0.8x + 0.5(0.8x) = 0.8x + 0.4x = 1.2x$.
To find the percentage of the original amount, we calculate $\frac{1.2x}{x} \times 100\% = 120\%$.
Thus, Brynn's savings are now 120% of the original amount.

Let's examine the choices:
A. 80: Too low because the final amount is more than the original.
B. 90: Too low for the same reason.
C. 100: Incorrect because the final amount is more than the original.
D. 110: Still too low.
E. 120: Matches our calculation.

ANSWER 1: E

Problem 2:
Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and an average of 44 pages per day for the next three days. She then finished the book by reading 10 pages on the last day. How many pages were in the book?

Let's calculate the total number of pages Isabella read.
In the first three days, she read $36 \times 3 = 108$ pages.
In the next three days, she read $44 \times 3 = 132$ pages.
On the last day, she read 10 pages.
So, the total number of pages in the book is $108 + 132 + 10 = 250$ pages.

Let's examine the choices:
A. 240: Too low because $108 + 132 + 10 = 250$, which is more than 240.
B. 250: Matches our calculation.
C. 260: Too high.
D. 270: Too high.
E. 280: Too high.

ANSWER 2: B

Problem 3:
Alice needs to replace a light bulb located 10 centimeters below the ceiling in her kitchen. The ceiling is 2.4 meters above the floor. Alice is 1.5 meters tall and can reach 46 centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

First, let's convert all measurements to centimeters.
The ceiling is $2.4 \times 100 = 240$ centimeters above the floor.
Alice's height is $1.5 \times 100 = 150$ centimeters.
The light bulb is $240 - 10 = 230$ centimeters above the floor.
Alice can reach $150 + 46 = 196$ centimeters without the stool.
So, the height of the stool is $230 - 196 = 34$ centimeters.

Let's examine the choices:
A. 32: Too low because $196 + 32 = 228$, which is less than 230.
B. 34: Matches our calculation.
C. 36: Too high.
D. 38: Too high.
E. 40: Too high.

ANSWER 3: B

Problem 4:
Billy's basketball team scored the following points over the course of the first 11 games of the season: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73. If his team scores 40 in the 12th game, which of the following statistics will show an increase?

Let's first arrange the scores in ascending order: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
The range is $73 - 42 = 31$. After adding 40, the new range is $73 - 40 = 33$, which is an increase.
The median is 58. After adding 40, the scores become 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73, and the median is the average of 58 and 58, which is still 58. So, the median doesn't change.
The mean is $\frac{42 + 47 + 53 + 53 + 58 + 58 + 58 + 61 + 64 + 65 + 73}{11} = \frac{632}{11} \approx 57.45$. After adding 40, the new mean is $\frac{632 + 40}{12} = \frac{672}{12} = 56$, which is a decrease.
The mode is 58. After adding 40, the mode is still 58. So, the mode doesn't change.
The mid-range is $\frac{73 + 42}{2} = 57.5$. After adding 40, the new mid-range is $\frac{73 + 40}{2} = 56.5$, which is a decrease.

Let's examine the choices:
A. range: Shows an increase.
B. median: Doesn't change.
C. mean: Decreases.
D. mode: Doesn't change.
E. mid-range: Decreases.

ANSWER 4: A

Problem 5:
Which of these numbers is less than its reciprocal?

Let's examine each choice:
A. $-2$: The reciprocal is $-\frac{1}{2}$. Since $-2 < -\frac{1}{2}$, $-2$ is less than its reciprocal.
B. $-1$: The reciprocal is $-1$. Since $-1 = -1$, $-1$ is not less than its reciprocal.
C. $0$: The reciprocal is undefined.
D. $1$: The reciprocal is $1$. Since $1 = 1$, $1$ is not less than its reciprocal.
E. $2$: The reciprocal is $\frac{1}{2}$. Since $2 > \frac{1}{2}$, $2$ is not less than its reciprocal.

Let's check if A is correct:
For $x = -2$, $x < \frac{1}{x}$ holds true because $-2 < -\frac{1}{2}$.

ANSWER 5: A

Problem 6:
There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Let's denote the number of boys as $x$ and the number of girls as $x + 4$.
The total number of students is $x + (x + 4) = 28$.
So, $2x + 4 = 28$, which simplifies to $2x = 24$, and $x = 12$.
Thus, the number of boys is 12 and the number of girls is $12 + 4 = 16$.
The ratio of girls to boys is $\frac{16}{12} = \frac{4}{3}$.

Let's examine the choices:
A. 3 : 4: Incorrect because the ratio of girls to boys is more than 1.
B. 4 : 3: Matches our calculation.
C. 3 : 2: Too high.
D. 7 : 4: Too high.
E. 2 : 1: Too high.

ANSWER 6: B

Problem 7:
In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

July has 31 days, and there are 24 hours in a day.
So, the total number of hours in July is $31 \times 24$.
The average rainfall per hour is $\frac{366}{31 \times 24}$.

Let's examine the choices:
A. $\frac{366}{31 \times 24}$: Matches our calculation.
B. $\frac{366 \times 31}{24}$: Too high.
C. $\frac{366 \times 24}{31}$: Too high.
D. $\frac{31 \times 24}{366}$: Too high and incorrect because it's the reciprocal.
E. $366 \times 31 \times 24$: Too high.

ANSWER 7: A

Problem 8:
The base of isosceles $\triangle ABC$ is 24 and its area is 60. What is the length of one of the congruent sides?

Let's denote the height of $\triangle ABC$ as $h$.
The area of $\triangle ABC$ is $\frac{1}{2} \times \text{base} \times h = 60$.
So, $\frac{1}{2} \times 24 \times h = 60$, which simplifies t

Problem 1:

Brynn’s savings first decrease by 20%, then increase by 50% of the new amount.

Let the original amount be 100.

- After a 20% decrease:  
  \[
  100-20=80
  \]
- Then the amount increases by 50% of 80:  
  \[
  50\%\text{ of }80=40
  \]
  So the new amount is:
  \[
  80+40=120
  \]

So Brynn’s savings are now 120% of the original amount.

Choices like 100% would be tempting if the 50% increase canceled the 20% decrease, but percentages are being taken from different amounts.

ANSWER 1: E

Problem 2:

Isabella reads for 7 days total.

First 3 days:

\[
3 \times 36=108
\]

Next 3 days:

\[
3 \times 44=132
\]

Last day:

\[
10
\]

Total pages:

\[
108+132+10=250
\]

So the book had 250 pages.

ANSWER 2: B

Problem 3:

The ceiling is 2.4 meters above the floor, which is:

\[
2.4\text{ m}=240\text{ cm}
\]

The light bulb is 10 cm below the ceiling, so its height above the floor is:

\[
240-10=230\text{ cm}
\]

Alice is 1.5 meters tall:

\[
1.5\text{ m}=150\text{ cm}
\]

She can reach 46 cm above her head, so without a stool she can reach:

\[
150+46=196\text{ cm}
\]

She needs to reach 230 cm, so the stool must add:

\[
230-196=34\text{ cm}
\]

ANSWER 3: B

Problem 4:

The original scores are:

\[
42,47,53,53,58,58,58,61,64,65,73
\]

There are 11 scores.

Original range:

\[
73-42=31
\]

Original median is the 6th score:

\[
58
\]

Original mode is 58.

Original mid-range:

\[
\frac{42+73}{2}=57.5
\]

Now add a 12th score of 40. The list becomes:

\[
40,42,47,53,53,58,58,58,61,64,65,73
\]

New range:

\[
73-40=33
\]

So the range increases.

New median is the average of the 6th and 7th scores:

\[
\frac{58+58}{2}=58
\]

So the median stays the same.

The mean decreases because 40 is below the old average.

The mode stays 58.

The new mid-range is:

\[
\frac{40+73}{2}=56.5
\]

So the mid-range decreases.

Only the range increases.

ANSWER 4: A

Problem 5:

We need a number that is less than its reciprocal.

Check each choice:

- For \(-2\), the reciprocal is \(-\frac12\). Since
  \[
  -2<-\frac12
  \]
  this works.
- For \(-1\), the reciprocal is \(-1\), so it is equal, not less.
- For \(0\), the reciprocal is undefined.
- For \(1\), the reciprocal is \(1\), so it is equal, not less.
- For \(2\), the reciprocal is \(\frac12\), and
  \[
  2>\frac12
  \]

Thus the correct choice is \(-2\).

ANSWER 5: A

Problem 6:

Let the number of boys be \(b\).

There are four more girls than boys, so the number of girls is:

\[
b+4
\]

There are 28 students total:

\[
b+(b+4)=28
\]

\[
2b+4=28
\]

\[
2b=24
\]

\[
b=12
\]

So there are 12 boys and:

\[
12+4=16
\]

girls.

The ratio of girls to boys is:

\[
16:12
\]

Simplify by dividing by 4:

\[
4:3
\]

ANSWER 6: B

Problem 7:

In July, there are 31 days.

Each day has 24 hours, so the total number of hours in July is:

\[
31 \times 24
\]

A total of 366 inches of rain fell, so the average rainfall per hour is:

\[
\frac{366}{31\times 24}
\]

This matches choice A.

ANSWER 7: A

Problem 8:

The base of the isosceles triangle is 24, and the area is 60.

Use the area formula:

\[
A=\frac12 bh
\]

\[
60=\frac12(24)h
\]

\[
60=12h
\]

\[
h=5
\]

In an isosceles triangle, the height to the base splits the base in half. So each half of the base is:

\[
\frac{24}{2}=12
\]

Now we have a right triangle with legs 5 and 12. The congruent side is the hypotenuse:

\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
\]

So one congruent side has length 13.

Choice A, 5, is just the height, not the side length.

ANSWER 8: C

Problem 9:

Use order of operations.

A:

\[
2+0+1+7=10
\]

B:

\[
2\times 0+1+7=0+1+7=8
\]

C:

\[
2+0\times 1+7=2+0+7=9
\]

D:

\[
2+0+1\times 7=2+0+7=9
\]

E:

\[
2\times 0\times 1\times 7=0
\]

The largest value is 10.

ANSWER 9: A

Problem 10:

The pitcher is \(\frac34\) full.

That amount is divided equally among 5 cups:

\[
\frac34 \div 5=\frac34 \times \frac15=\frac{3}{20}
\]

Each cup receives \(\frac{3}{20}\) of the total pitcher capacity.

Convert to a percent:

\[
\frac{3}{20}=0.15=15\%
\]

ANSWER 10: C

Problem 11:

The expression is:

\[
\frac{\frac38+\frac78}{\frac45}
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54
\]

Now divide by \(\frac45\):

\[
\frac54 \div \frac45
\]

Dividing by a fraction means multiplying by its reciprocal:

\[
\frac54 \times \frac54=\frac{25}{16}
\]

So the value is:

\[
\frac{25}{16}
\]

ANSWER 11: B

Problem 12:

Let the number be \(x\).

We are told that 20% of the number is 12:

\[
0.20x=12
\]

Since \(0.20=\frac15\), this means:

\[
\frac15x=12
\]

So:

\[
x=60
\]

Now find 30% of 60:

\[
0.30 \times 60=18
\]

ANSWER 12: B

Problem 1:
We are asked to find what percent of her original savings Brynn has after a 20% decrease followed by a 50% increase on the new amount.

Let Brynn's original savings be $100. 
1. In July, her savings decreased by 20%. 
   Decrease = 20% of $100 = $20.
   Savings after July = $100 - $20 = $80.
2. In August, her savings increased by 50% of the *new* amount.
   Increase = 50% of $80 = $40.
   Savings after August = $80 + $40 = $120.

Her final savings are $120, which is exactly 120% of her original $100. This rules out 80, 90, 100, and 110.

ANSWER 1: E

Problem 2:
We need to find the total number of pages in the book Isabella read over a week.

We can break down her reading into three parts and sum them up:
1. First 3 days: She read an average of 36 pages per day.
   Pages read = 36 × 3 = 108 pages.
2. Next 3 days: She read an average of 44 pages per day.
   Pages read = 44 × 3 = 132 pages.
3. Last day (7th day): She read 10 pages.

Total pages = 108 + 132 + 10 = 240 + 10 = 250 pages.
This rules out 240, 260, 270, and 280.

ANSWER 2: B

Problem 3:
We are asked to find the height of the stool Alice needs to stand on to reach a light bulb.

First, let's convert all measurements to centimeters to keep the units consistent:
- The ceiling is 2.4 meters = 240 cm above the floor.
- The light bulb is 10 cm below the ceiling, so its height from the floor is 240 - 10 = 230 cm.
- Alice is 1.5 meters = 150 cm tall.
- She can reach 46 cm above her head, so her maximum reach from the floor while standing flat is 150 + 46 = 196 cm.

To reach the light bulb at 230 cm, the stool must make up the difference:
Height of stool = Height of light bulb - Alice's reach
Height of stool = 230 cm - 196 cm = 34 cm.
This rules out 32, 36, 38, and 40.

ANSWER 3: B

Problem 4:
We need to determine which statistic of the team's scores increases when a 12th game score of 40 is added to the data set.

Original 11 games: 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: Maximum - Minimum = 73 - 42 = 31.
- Median: The middle (6th) value is 58.
- Mean: Sum / 11 = 632 / 11 ≈ 57.45.
- Mode: The most frequent value is 58.
- Mid-range: (Maximum + Minimum) / 2 = (73 + 42) / 2 = 57.5.

New 12 games (adding 40): 40, 42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73.
- Range: 73 - 40 = 33. (This has increased from 31).
- Median: The average of the 6th and 7th values = (58 + 58) / 2 = 58. (No change).
- Mean: The new score (40) is lower than the original mean (57.45), so the mean will decrease.
- Mode: 58 is still the most frequent. (No change).
- Mid-range: (73 + 40) / 2 = 56.5. (Decreased).

The only statistic that shows an increase is the range.

ANSWER 4: A

Problem 5:
We are asked to find which of the given numbers is strictly less than its reciprocal.

Let's test each choice:
A. −2: The reciprocal is −1/2. Since −2 is further to the left on the number line than −0.5, −2 < −1/2 is True.
B. −1: The reciprocal is −1. −1 < −1 is False.
C. 0: The reciprocal is undefined (division by zero).
D. 1: The reciprocal is 1. 1 < 1 is False.
E. 2: The reciprocal is 1/2. 2 < 1/2 is False.

ANSWER 5: A

Problem 6:
We need to find the ratio of girls to boys in a class of 28 students where there are 4 more girls than boys.

Let $b$ be the number of boys.
The number of girls is $g = b + 4$.
The total number of students is $b + g = 28$.
Substitute $g$:
$b + (b + 4) = 28$
$2b + 4 = 28$
$2b = 24$
$b = 12$

Now find the number of girls:
$g = 12 + 4 = 16$.

The ratio of the number of girls to the number of boys is:
$g : b = 16 : 12$
Dividing both sides by 4, we get $4 : 3$.
This rules out 3:4, 3:2, 7:4, and 2:1.

ANSWER 6: B

Problem 7:
We need to find the average rainfall in inches per hour during the month of July.

- Total rainfall for the month = 366 inches.
- Number of days in July = 31 days.
- Number of hours in a day = 24 hours.
- Total hours in July = 31 × 24.

To find the average rainfall per hour, we divide the total rainfall by the total number of hours:
Average rainfall = 366 / (31 × 24).
This matches choice A exactly.

ANSWER 7: A

Problem 8:
We are asked to find the length of one of the congruent sides of an isosceles triangle given its base and area.

Let the base $b = 24$ and the height to the base be $h$.
The formula for the area of a triangle is:
Area = (1/2) × base × height
60 = (1/2) × 24 × $h$
60 = 12$h$
$h = 5$

In an isosceles triangle, the altitude to the base bisects the base. This creates two identical right-angled triangles, each with:
- One leg as the height ($h = 5$)
- The other leg as half the base (24 / 2 = 12)
- The hypotenuse as the congruent side ($s$)

Using the Pythagorean theorem:
$s^2 = 5^2 + 12^2$
$s^2 = 25 + 144$
$s^2 = 169$
$s = 13$
This rules out 5, 8, 14, and 18.

ANSWER 8: C

Problem 9:
We need to evaluate each expression using the standard order of operations (PEMDAS/BODMAS) and find the largest value.

A. 2 + 0 + 1 + 7 = 10
B. 2 × 0 + 1 + 7 = 0 + 1 + 7 = 8
C. 2 + 0 × 1 + 7 = 2 + 0 + 7 = 9
D. 2 + 0 + 1 × 7 = 2 + 0 + 7 = 9
E. 2 × 0 × 1 × 7 = 0

Comparing the results (10, 8, 9, 9, 0), the largest value is 10.

ANSWER 9: A

Problem 10:
We are asked to find what percent of the pitcher's total capacity is poured into each of the 5 cups.

The pitcher initially contains juice equal to 3/4 of its total capacity.
This amount is divided equally among 5 cups.
Amount per cup = (3/4) / 5 = 3/20 of the pitcher's total capacity.

To convert this fraction to a percentage, we multiply by 100:
(3/20) × 100% = 3 × 5% = 15%.
This rules out 5, 10, 20, and 25.

ANSWER 10: C

Problem 11:
The text "3 8 + 7 8 4 5 =" represents the complex fraction problem:
(3/8 + 7/8) / (4/5)

Let's evaluate the numerator first:
3/8 + 7/8 = 10/8 
We can simplify 10/8 to 5/4.

Now, we divide this result by the denominator:
(5/4) / (4/5)

To divide by a fraction, we multiply by its reciprocal:
(5/4) × (5/4) = 25/16.
This matches choice B perfectly and rules out 1, 2, 43/20, and 47/16.

ANSWER 11: B

Problem 12:
We are asked to find 30% of a number