Full Length GRE Quantitative Reasoning Practice Test

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GRE Quantitative Reasoning Practice Test 1

Section 1 20 questions Total time for this section: 35 Minutes You can use a basic calculator on this section.

$a$ and

$b$ are real numbers.

$a<b$

Quantity A	Quantity B
$\|a\ -\ b\|$	$\|b\ -\ \|$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

$6$ percent of

$x$ is equal to

$5$ percent of

$y$ , where

$x$ and

$y$ are positive numbers.

Quantity A	Quantity B
$x$	$y$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

$x$ is an integer greater than zero.

Quantity A	Quantity B
$\frac{ 1}{x}\ +\ x$	$8$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

Quantity A	Quantity B
The least prime factor of $55$	The least prime factor of $210$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

Emma and Sophia have a family business. The profit of their business will be divided between Emma and Sophia in the ratio

$3$ to

$4$ respectively.;

Quantity A	Quantity B
The money Emma receives when the profit is $$560$ .	The money Sophia receives when the profit is $$420$ .

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

The volume of a sphere with diameter of length 5 is how many times the volume of sphere with diameter of length

$\sqrt{5}$ ?
(Volume of a sphere

$=\frac{ 4}{3} \ π\ r^3)$ .

(A)

$\sqrt{5}$

(B)

${5}$

(C)

$10$

(D)

$10\sqrt{5}$

(E)

$5\sqrt{5}$

What is the solution of the following system of equations?

$\begin{cases}-\ \frac{x}{2}\ +\ \frac{y}{4}=1\\ -\ \frac{5\ y} {6}\ +\ 2\ x=4\end{cases}$

(A)

$x=48, \ y=22$

(B)

$x=50, \ y=20$

(C)

$x=20, \ y=50$

(D)

$x=22, \ y=48$

(E)

$x=22, \ y=50$

$y=2^4$ then what is the value of

$y^\sqrt{y}$ ?

(A)

$2$

(B)

$2^4$

(C)

$2^8$

(D)

$2^{16}$

(E)

$2^{32}$

The average weight of

$18$ girls in a class is

$60$ kg and the average weight of

$32$ boys in the same class is

$62$ kg. What is the average weight of all the

$50$ students in that class?

(A)

$60$

(B)

$61.28$

(C)

$61.68$

(D)

$61.90$

(E)

$62.20$

10-

$60 \%$ of A is

$20 \%$ of B, then B is what percent of A?

(A)

$3\%$

(B)

$30\%$

(C)

$200\%$

(D)

$300\%$

(E)

$900\%$

11-

$(x\ -\ 2)^3=27$ which of the following could be the value of

$(x\ -\ 4)\ (x\ -\ 3)$ ?

(A)

$1$

(B)

$2$

(C)

$6$

(D)

$-\ 1$

(E)

$-\ 2$

12-

The average of

$x , y$ and

$5$ is

$5$ and

$x\ -\ y=-\ 8$ What is the value of

$x\ ×\ y$ ?

(A)

$9$

(B)

$- \ 9$

(C)

$- \ 8$

(D)

$8$

(E)

$0$

13-

The surface area of a cylinder is

$150\ π$ cm

$^2$ . If its height is

$10$ cm, what is the radius of the cylinder?

(A)

$13$ cm

(B)

$11$ cm

(C)

$15$ cm

(D)

$5$ cm

(E)

$7$ cm

14-

In the

$xy$ -plane, the point

$(4,3)$ and

$(3,2)$ are on line A. Which of the following equations of lines is parallel to line A?

(A)

$y= 3\ x$

(B)

$y= 10$

(C)

$y= \frac{x}{2}$

(D)

$y=2\ x$

(E)

$y=x$

15-

What is the product of the number of Mathematics and number of English books?
GRE_Quantitative

(A)

$21,168$

(B)

$31,752$

(C)

$26,460$

(D)

$17,640$

(E)

$14,112$

16-

What are the values of angle

$α$ and

$β$ ?

(A)

$90^ \circ , 54^\circ$

(B)

$120^ \circ , 36^\circ$

(C)

$120^ \circ , 45^\circ$

(D)

$108^ \circ , 54^\circ$

(E)

$108^ \circ , 36^\circ$

17-

The librarians decided to move some of the books in the Mathematics section to Chemistry section. How many books are in the Chemistry section if now

$γ=\frac{2}{5}\ α$ ?
GRE_Quantitative2

(A)

$80$

(B)

$120$

(C)

$150$

(D)

$180$

(E)

$200$

18-

Let

$r$ and p be constants. If

$x^2\ +\ 6\ x\ +\ r$ factors into

$(x \ +\ 2)\ (x \ +\ p)$ , the values of

$r$ and

$p$ respectively are?

(A)

$8, 4$

(B)

$4, 8$

(C)

$6, 3$

(D)

$3, 6$

(E)

The answer cannot be found from the information given.

19-

In how many ways can

$5$ cards be placed in

$3$ positions if any cards can be placed in any position?

(A)

$5$

(B)

$10$

(C)

$15$

(D)

$30$

(E)

$120$

20-

$(x\ -\ 2)^2\ +\ 1\ >\ 3\ x\ -\ 1$ , then

$x$ can equal to which of the following?

(A)

$1$

(B)

$6$

(C)

$8$

(D)

$3$

(E)

$4$

GRE Quantitative Reasoning Practice Test 1

Section 2 20 questions Total time for this section: 35 Minutes You can use a basic calculator on this section.

21-

Quantity A	Quantity B
$(-\ 5)^4$	$5^4$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

22-

The average of

$3, 4$ , and

$x$ is

$3$ .

Quantity A	Quantity B
$x$	average of $x,\ x\ -\ 6,\ x\ +\ 4,\ 2\ x$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

23-

$\frac{4}{5}<\ x\ <\frac{6}{7}$

Quantity A	Quantity B
$x$	$\frac{5}{6}$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

24-

$n$ is a natural number and

$\frac{1}{3^n} <\frac{1}{27}$

Quantity A	Quantity B
$3$	$n$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

25-

Quantity A	Quantity B
$\frac{ x^6}{6}$	$(\frac{x}{6})^6$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

26-

Quantity A	Quantity B
$5\ +\ 8\ ×\ (–\ 2)\ –\ [4\ +\ 22\ ×\ 5]\ ÷\ 6$	$[6\ ×\ (–\ 24)\ +\ 8]\ –\ (–\ 4)\ +\ [4\ ×\ 5]\ ÷\ 2$

(A)

Quantity A is greater.

(B)

Quantity B is greater.

(C)

The two quantities are equal.

(D)

The relationship cannot be determined from the information given.

27-

$|a|<1$ , then which of the following is true?

$(b>0)$ ?
I.

$–\ b<b\ a <b$
II.

$-\ a<a^2<a$ if

$a<0$
III.

$-\ 5<2\ a\ -\ 3<-\ 1$

(A)

I only

(B)

II only

(C)

I and III only

(D)

III only

(E)

I, II and III

28-

The ratio of boys and girls in a class is

$4:7$ . If there are

$44$ students in the class, how many more boys should be enrolled to make the ratio

$1:1$ ?

(A)

$8$

(B)

$10$

(C)

$12$

(D)

$14$

(E)

$28$

29-

$150 \%$ of a number is

$75$ , then what is

$90 \%$ of that number?

(A)

$45$

(B)

$50$

(C)

$70$

(D)

$85$

(E)

$90$

30-

Removing which of the following numbers will change the average of the numbers to

$6$ ?

$1, \ 4, \ 5, \ 8, \ 11, \ 12$

(A)

$1$

(B)

$4$

(C)

$5$

(D)

$11$

(E)

$12$

31-

The length of a rectangle is

$\frac{5}{4}$ times its width. If the width is

$16$ , what is the perimeter of this rectangle?

(A)

$36$

(B)

$48$

(C)

$72$

(D)

$144$

(E)

$180$

32-

The marked price of a computer is D dollar. Its price decreased by

$20\%$ in January and later increased by

$10 \%$ in February. What is the final price of the computer in D dollar?

(A)

$0.80$ D

(B)

$0.88$ D

(C)

$0.90$ D

(D)

$1.20$ D

(E)

$1.40$ D

33-

In the following figure, what is the perimeter of

$∆ ABC$ if the area of

$∆ ADC$ is

$15$ ?

(A)

$37.5$

(B)

$21$

(C)

$15$

(D)

$24$

(E)

The answer cannot be determined from the information given

34-

A line

$l$ is parallel to the

$x$ -axis and passes through the point

$(-3,4)$ . What is the slope of the line

$(m)$ and its

$y$ -intercept?

(A)

$𝑚=∞, y\ −$ intercept

$=4$

(B)

$𝑚=∞, y\ −$ intercept

$=-\ 3$

(C)

$𝑚=0, y\ −$ intercept

$=-\ 3$

(D)

$𝑚=0, y\ −$ intercept

$=4$

(E)

$𝑚=-\ 3, y\ −$ intercept

$=4$

35-

Between which two of the months shown was there a twenty percent decreased in the number of pants sold?
GRE_Quantitative4

(A)

January and February

(B)

February and March

(C)

March and April

(D)

April and May

(E)

May and June

36-

During the six-month period shown, what is the median number of shirts and mean number of shoes per month?
GRE_Quantitative5

(A)

$146.5, 30$

(B)

$147.5, 29$

(C)

$146.5, 31$

(D)

$147.5, 30$

(E)

$146.5, 29$

37-

How many shoes need to be added in April until the ratio of number of pants to number of shoes in April equals to five-seventeenth of this ratio in May?
GRE_Quantitative6

(A)

$90$

(B)

$80$

(C)

$70$

(D)

$60$

(E)

$50$

38-

What is the product of all possible values of

$x$ in the following equation?

$|x\ -\ 2\ x\ -\ 5\ +\ 7|=4$

(A)

$12$

(B)

$-\ 12$

(C)

$6$

(D)

$-\ 6$

(E)

$0$

39-

In the following figure, ABCD is a rectangle, and E and F are points on AD and DC, respectively and DE

$=4$ and DF

$=3$ . The area of ∆BED is

$16$ , and the area of ∆BDF is

$18$ . What is the perimeter of the rectangle?
GRE_Quantitative7

(A)

$20$

(B)

$22$

(C)

$32$

(D)

$40$

(E)

$44$

40-

In the following figure, ABCD is a rectangle. If a

$=\sqrt{3}$ , and b

$=2\ a$ , find the area of the shaded region. (the shaded region is a trapezoid)
GRE_Quantitative8

(A)

$4$

(B)

$2$

(C)

$\sqrt{3}$

(D)

$2\sqrt{3}$

(E)

$4\sqrt{3}$

GRE Quantitative Reasoning Practice Test Answers and Explanations

1- Choice C is correct

The correct answer is The two quantities are equal.
Choose different values for a and b and find the values of quantity A and quantity B. a $=2$ and b $=3$ , then:
Quantity A: $|2\ -\ 3|=|-\ 1|=1$
Quantity B: $|3\ -\ 2|=|1|=1$ The two quantities are equal. a $=-\ 3$ and b $=2$ , then:
Quantity A: $|-\ 3\ -\ 2|=|-\ 5|=5$ Quantity B: $|2\ -\ (-\ 3)|=|2\ +\ 3|=5$ The two quantities are equal. Any other values of a and b provide the same answer.

2- Choice B is correct

The correct answer is Quantity B is greater.
$6\%$ of $= 5\%$ of $y → 0.06\ x = 0.05\ y→x=\frac{0.05}{0.06} \ y→x=\frac{5}{6}\ y$ , therefore, $y$ is bigger than $x$ .

3- Choice D is correct

The correct answer is The relationship cannot be determined from the information given.
Choose different values for x and find the value of quantity A.
$x=1$ ,
then: Quantity A: $\frac{1}{x}\ +\ x= \frac{ 1}{1}\ +\ 1=2$
Quantity B is greater $x=0.1$ , then:
Quantity A: $\frac{1}{x}\ +\ x= \frac{ 1}{0.1}\ +\ 1=10\ +\ 1=11$
Quantity A is greater
The relationship cannot be determined from the information given.

4- Choice A is correct

The correct answer is Quantity A is greater.
prime factoring of $55$ is: $5\ ×\ 11$
prime factoring of $210$ is: $2\ ×\ 3\ ×\ 5\ ×\ 7$
Quantity A $= 5$ and Quantity B $= 2$

5- Choice C is correct

The correct answer is The two quantities are equal.
The profit of their business will be divided between Emma and Sophia in the ratio $3$ to $4$ respectively.
Therefore, Emma receives $\frac{3}{7}$ of the whole profile and Sophia receives $\frac{4}{7}$ of the whole profile.
Quantity A: The money Emma receives when the profit is $$560$ equals: $\frac{3}{7}\ ×\ 560=240$
Quantity B: The money Sophia receives when the profit is $$420$ equals: $\frac{4}{7}\ ×\ 420=240$
The two quantities are equal.

6- Choice E is correct

The correct answer is $5\sqrt{5}$
$V_1=\frac{4\ π}{3}\ (\frac{5}{2})^3$
$V_2=\frac{4\ π}{3}\ (\frac{\sqrt5 }{2})^3 → \frac{ V_1}{V_2} =5\sqrt5$

7- Choice D is correct

The correct answer is $x=22, \ y=48$
$\begin{cases}-\ \frac{x}{2}\ +\ \frac{y}{4}=1\\ -\ \frac{5\ y} {6}\ +\ 2\ x=4\end{cases}$ →multiply the top equation by 4 then:
$\begin{cases} -\ 2\ x\ +\ y=4 \\ -\ \frac{5\ y}{6}\ +\ 2\ x\ =4 \end{cases}$ →add two equations
$\frac{1}{6} \ y=8→y=48$ , plug in the value of y into the first euation. $→ x=22$

8- Choice D is correct

The correct answer is $2^{16}$
We know that, $\sqrt[n]{a^m }=a^{\frac{m}{n}}$ then:
$\sqrt{y}=\sqrt{2^4} =2^2=4→(2^4)^4=2^{16}$

9- Choice B is correct

The correct answer is $61.28$
average $= \frac{sum \ of\ terms }{number \ of\ terms}$
The sum of the weight of all girls is: $18 \ ×\ 60 = 1080$ kg
The sum of the weight of all boys is: $32 \ ×\ 62 = 1984$ kg
The sum of the weight of all students is: $1080 \ +\ 1984 = 3064$ kg
average $= \frac{3064 }{50} = 61.28$

10- Choice D is correct

The corrcet answer is $300\%$
Write the equation and solve for B: $0.60$ A $= 0.20$ B, divide both sides by $0.20$ , then you will have
$\frac{0.60}{0.20}$ A = B, therefore: B $= 3$ A, and B is $3$ times of A or it’s $300\%$ of A.

11- Choice B is correct

The correct answer is $2$
$(x\ -\ 2)^3=27→x\ -\ 2=3→x=5 →(x\ -\ 4)\ (x\ -\ 3)=(5\ -\ 4)\ (5\ -\ 3)=(1)\ (2)=2$

12- Choice A is correct

The correct answer is $9$
average $=\frac{sum \ of\ terms}{number \ of\ terms} → \frac{x\ +\ y\ +\ 5}{3}=5→x\ +\ y=10$
$\begin{cases}x\ +\ y=10\\ x\ -\ y=-\ 8\end{cases}$ add both equations: $2\ x=2→x=1→y=9→x\ ×\ y=9$

13- Choice D is correct

The correct answer is $5$ cm
Formula for the Surface area of a cylinder is:
$SA=2\ π\ r^2+2\ π\ r\ h\ \to 150\ π=2\ π\ r^2\ +\ 2\ π\ r\ (10)\to r^2\ +\ 10\ r\ -\ 75=0$
$(r\ +\ 15)\ (r\ -\ 5)=0\to r=5$ or $r= -\ 15$ (unacceptable)

14- Choice E is correct

The correct answer is $y = x$
The slop of line A is: $m=\frac{y_2\ -\ y_1}{x_2\ -\ x_1 }=\frac{3\ -\ 2}{4\ -\ 3}=1$
Parallel lines have the same slope and only choice E $(y=)$ has slope of $1$ .

15- Choice B is correct

The correct answer is $31,752$
number of Mathematics book: $0.3\ ×\ 840=252$
number of English book: $0.15\ ×\ 840=126$
product of number of Mathematics and number of English book: $252\ ×\ 126=31,752$

16- Choice D is correct

The correct answer is $108^°,54^°$
The angle $α$ is: $0.3\ ×\ 360=108^°$
The angle $β$ is: $0.15\ ×\ 360=54^°$

17- Choice B is correct

The correct answer is $120$
According to the chart, $50\%$ of the books are in the Mathematics and Chemistry sections.
Therefore, there are $420$ books in these two sections.
$0.50 \ ×\ 840 = 420$
$γ\ +\ α=420$ , and $γ=\frac{2}{5}\ α$
Replace $γ$ by $\frac{2}{5} \ α$ in the first equation.
$γ\ +\ α=420→\frac{2}{5} α\ +\ α=420→\frac{7}{5} \ α=420→$ multiply both sides by $\frac{5}{7}$
$(\frac{5}{7})\ \frac{7}{5} α=420\ ×\ (\frac{5}{7})→α=\frac{420\ ×\ 5}{7}=300$
$α=300→γ=\frac{2}{5} α→γ=\frac{2}{5}\ ×\ 300=120$
There are $120$ books in the Chemistry section.

18- Choice A is correct

The correct answer is $8$
We have: $(x\ +\ 2)\ (x\ +\ p)=x^2\ +\ (2\ +\ p)\ x\ +\ 2\ p→2\ +\ p=6→p=4$ and $r=2\ p=8$

19- Choice B is correct

The correct answer is $10$
This question is a combination problem. The formula for combination is:
nCr $= \frac{n!}{r!\ (n\ -\ r)!}$
This formula is for the number of possible combinations of $r$ objects from a set of $n$ objects.
Using the information in the question:
$5$ C $3 = \frac{5!}{3!\ (5\ -\ 3)!}=\frac{5\ × \ 4\ ×\ 3\ ×\ 2\ ×\ 1}{3\ ×\ 2\ ×\ 1\ ×\ (2\ ×\ 1)}=10$

20- Choice C is correct

The correct answer is $8$
Plug in the value of each option in the inequality.
A. $1$ $(1\ -\ 2)^2\ +\ 1>3\ (1)\ -\ 1→2>2$ No!
B. $6$ $(6\ -\ 2)^2\ +\ 1>3\ (6)\ -\ 1→17>17$ No!
C. $8$ $(8\ -\ 2)^2\ +\ 1>3\ (8)\ -\ 1→37>23$ Bingo!
D. $3$ $(3 \ -\ 2)^2\ +\ 1 >3\ (3)\ -\ 1→2>8$ No!
C $4$ $(4\ -\ 2)^2\ +\ 1>3\ (4)\ -\ 1→5>11$ No!

20- Choice C is correct

21- Choice C is correct

The correct answer is The two quantities are equal.
Simplify both quantities.
Quantity A: $(-\ 5)^4=(-\ 5)\ ×\ (-\ 5)\ ×\ (-\ 5)\ × \ (- \ 5)=625$
Quantity B: $5\ ×\ 5\ ×\ 5\ ×\ 5=625$
The two quantities are equal.

22- Choice C is correct

The correct answer is The two quantities are equal.
Quantity A is: $\frac{3\ +\ 4\ +\ x}{3}=3→x=2$
Quantity B is: $\frac{2\ +\ (2\ -\ 6)\ +\ (2\ +\ 4)\ +\ (2\ ×\ 2)}{4}=2$

23- Choice D is correct

The correct answer is The relationship cannot be determined from the information given.
Simply change the fractions to decimals.
$\frac{4}{5}=0.80$
$\frac{6}{7}=0.857…$
$\frac{5}{6}=0.8333…$
As you can see, $x$ lies between $0.80$ and $0.857…$ and it can be $0.81$ or $0.84$ .
The first one is less than $0.833…$ and the second one is greater than $0.833…$
The relationship cannot be determined from the information given.

24- Choice B is correct

The correct answer is Quantity B is greater.
$\frac{1}{3^n} <\frac{1}{27} → 3^{-\ n}<3^{-\ 3}→-\ <-\ 3$ ,divide both side by $-\ 1→n>3$

25- Choice A is correct

The correct answer is Quantity A is greater.
Simplify quantity B.
Quantity B: $(\frac{x}{6})^6=\frac{x^6}{6^6}$
Since, the two quantities have the same numerator $(x^6)$ and the denominator in quantity B is bigger $(6^6>6)$ , then the quantity A is greater.

26- Choice A is correct

The correct answer is Quantity A is greater.
Use PEMDAS (order of operation):
Quantity A $= 5 \ +\ 8 \ ×\ (–\ 2)\ –\ [4 \ +\ 22 \ ×\ 5] \ ÷\ 6 = 5 \ +\ 8 \ ×\ (–\ 2) \ –\ [4 \ +\ 110] \ ÷\ 6 =$
$5 \ +\ 8 \ ×\ (–\ 2) \ –\ [114] \ ÷\ 6 = 5 \ +\ (–\ 16) \ –\ 19 = 5 \ +\ (–\ 16) \ – \ 19 = –\ 11 \ –\ 19 = –\ 30$
Quantity B $= [6 \ ×\ (–\ 24) \ +\ 8] \ –\ (–\ 4)\ +\ [4 \ ×\ 5] \ ÷\ 2 = [–\ 144 \ +\ 8] \ –\ (–\ 4) \ +\ [20] \ ÷\ 2 =$
$[–\ 144 \ +\ 8] \ –\ (–\ 4) \ +\ 10 = [–\ 136] \ –\ (–\ 4) \ +\ 10 = [–\ 136] \ +\ 4 \ +\ 10 = –\ 122$
$-\ 30>-\ 122$

27- Choice C is correct

The correct answer is I and III only.
I. $|a|<1→-\ 1<a<1$
Multiply all sides by b. Since, $b>0→-\ b<b\ a<b$
II. Since, $-\ 1<a<1$ ,and $a<0→-\ a>a^2>a$ (plug in $\frac{-\ 1}{2}$ , and check!)
III. $-\ 1<a<1$ ,multiply all sides by 2,then: $-\ 2<2a<2$ ,subtract 3 from all sides,the:
$-\ 2\ -\ 3<2\ a\ -\ 3<2\ -\ 3→-\ 5<2\ a\ -\ 3<-\ 1$

28- Choice C is correct

The correct answer is $12$
The ratio of boy to girls is $4:7$ . Therefore, there are $4$ boys out of $11$ students.
To find the answer, first divide the total number of students by $11$ , then multiply the result by $4$ .
$44 \ ÷\ 11 = 4 ⇒ 4 \ ×\ 4 = 16$
There are $16$ boys and $28\ (44 \ –\ 16)$ girls. So, $12$ more boys should be enrolled to make the ratio $1:1$

29- Choice A is correct

The correct answer is $45$
First, find the number.
Let $x$ be the number. Write the equation and solve for $x$ .
$150 \%$ of a number is $75$ , then:
$1.5\ ×\ x=75 ⇒ x=75\ ÷\ 1.5=50$
$90 \%$ of $50$ is:
$0.9 \ ×\ 50 = 45$

30- Choice D is correct

The correct answer is $11$
Check each option provided:
A. $1$ $\frac{4\ +\ 5\ +\ 8\ +\ 11\ +\ 12}{5}=\frac{40}{5}=8$
B. $4$ $\frac{1\ +\ 5\ +\ 8\ +\ 11\ +\ 12}{5}=\frac{37}{5}=7.4$
C. $5$ $\frac{1\ +\ 4\ +\ 8\ +\ 11\ +\ 12}{5}=\frac{36}{5}=7.2$
D. $11$ $\frac{1\ +\ 4\ +\ 5\ +\ 8 \ +\ 12}{5}=\frac{30}{5}=6$
E. $12$ $\frac{1\ +\ 4\ +\ 5\ +\ 8\ +\ 11}{5}=\frac{29}{5}=5.8$

31- Choice C is correct

The correct answer is $72$
length of the rectangle is: $\frac{5}{4}\ ×\ 16=20$
perimeter of rectangle is: $2\ ×\ (20\ +\ 16)=72$

32- Choice B is correct

The correct answer is $0.88$ D
To find the discount, multiply the number by $(100\% \ –$ rate of discount).
Therefore, for the first discount we get: (D) $(100\% \ –\ 20\%) =$ (D) $(0.80) = 0.80$ D
For increase of $10\%: (0.85$ D) $(100\% \ +\ 10\%) = (0.85$ D) $(1.10) = 0.88$ D $= 88\%$ of D

33- Choice D is correct

The correct answer is $24$
Let $x$ be the length of AB, then:
$15=\frac{x\ ×\ 5}{2}→x=6$
The length of AC $=\sqrt{6^2\ +\ 8^2 }=\sqrt{100}=10$
The perimeter of ∆ABC $=6\ +\ 8\ +\ 10=24$

34- Choice D is correct

The correct answer is $𝑚=0, y\ −$ intercept $=4$
Since line $l$ is parallel to $x$ -axis, therefore the slope of $l$ is equal to $0$ and the value of $y$ is the same as the value of $y$ in the point $(-\ 3, 4)$ . Therefore, $y$ -intercept is $4$ .

35- Choice A is correct

The correct answer is January and February
First find the number of pants sold in each month.
A. January: $110$ , February: $88$ , March: $90$ , April: $70$ , May: $85$ , June: $65$
Check each option provided.
January and February,
$(\frac{110 \ - \ 88}{110}) \ × \ 100=\frac{22}{110} \ × \ 100=20\%$
B. February and March, there is an increase from February to March.
C. March and April
$(\frac{90 \ - \ 70}{90}) \ × \ 100=\frac{20}{90} \ × \ 100=22.22\%$
D. April and May: there is an increase from April to May
May and June
$(\frac{85 \ - \ 65}{85}) \ × \ 100=\frac{20}{85} \ × \ 100=23.53\%$

36- Choice D is correct

The correct answer is $147.5, 30$
Let’s order number of shirts sold per month:
$130,140,145,150,160,170$
median is: $\frac{145 \ + \ 150}{2}=147.5$
Let’s list the number of shoes sold per month:
$20,25,25,35,35,40$
mean is: $\frac{20 \ + \ 25 \ + \ 25 \ + \ 35 \ + \ 35 \ + \ 40}{6}=\frac{180}{6}=30$

37- Choice E is correct

The correct answer is $50$
Let $x$ be the number of shoes need to be added in April. Then:
$\frac{70}{20 \ + \ x}=(\frac{5}{17}) \ (\frac{85}{25}) →\frac{70}{20 \ + \ x}=\frac{425}{425}=1→$
$70=20 \ + \ x→x=50$

38- Choice B is correct

The correct answer is $-\ 12$
$|x\ -\ 2\ x\ -\ 5\ +\ 7|=4→ |-\ x\ +\ 2|=4 →-\ x\ +\ 2=4$ or $-\ x\ +\ 2=-\ 4$
$→x=-\ 2$ or $x=6$
The product of all possible values of $x = (-\ 2)\ ×\ 6=-\ 12$

39- Choice D is correct

The correct answer is $40$
The area of ∆BED is $16$ , then: $\frac{4\ ×\ AB}{2}=16→4\ ×\ AB=32→AB=8$
The area of ∆BDF is $18$ , then: $\frac{3\ ×\ BC}{2}=18→3\ ×\ BC=36→BC=12$
The perimeter of the rectangle is $= 2\ ×\ (8\ +\ 12)=40$

40- Choice E is correct

The correct answer is $4\sqrt{3}$
Based on triangle similarity theorem:
$\frac{a}{a\ +\ b}=\frac{c}{3}→c=\frac{3\ a}{a\ +\ b}=\frac{3\sqrt3}{3\sqrt3}=1→$
area of shaded region is:
$(\frac{c\ +\ 3}{2})\ (b)=\frac{4}{2}\ ×\ 2\sqrt{3}=4\sqrt{3}$

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