Full Length TASC Mathematics Practice Test

The best way to prepare for the TASC math test is by taking a practice test. Not only will this simulate what it would be like on exam day, but this will help you feel more confident and measure your readiness to take the actual exam.

In order to get the most out of this practice test and prepare your mind, body, and spirit for the actual TASC Math test (which is also a realistic resource), we recommend you treat it as if it were an actual one. Clear away any distractions with scratch paper in hand, pencil ready to go, timer ticking down every second as well as calculator on standby. Take this in one sitting so you can quickly assess your score at the end!

Take this practice test to simulate the experience of taking a full-length TASC Math Test Day. After you've finished, use the answer keys to score your tests. Best of luck!

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TASC Mathematics Practice Test 4 Section 1 (Calculator) 40 questions Total time for this section: 50 Minutes You may use a calculator on this Section.
1-	What is the volume of a box with the following dimensions? Height $= 3$ cm Width $= 5$ cm Length $= 8$ cm
(A)	$120$ cm$^3$
(B)	$130$ cm$^3$
(C)	$142$ cm$^3$
(D)	$112$ cm$^3$
2-	Which of the following points lies on the line $x \ + \ 3 \ y=8$?
(A)	$(− \ 2, 3)$
(B)	$(2, 2)$
(C)	$(− \ 2, 3)$
(D)	$(− \ 3, 4)$
3-	Last week $20,000$ fans attended a football match. This week three times as many bought tickets, but one fifth of them cancelled their tickets. How many are attending this week?
(A)	$45,000$
(B)	$38,000$
(C)	$48,000$
(D)	$42,000$
4-	Two dice are thrown simultaneously, what is the probability of getting a sum of $6$ or $9$?
(A)	$\frac{1}{2}$
(B)	$\frac{2}{5}$
(C)	$\frac{1}{4}$
(D)	$\frac{1}{3}$
5-	What is the value of $4^5$?
(A)	$1,024$
(B)	$1,125$
(C)	$1,140$
(D)	$1,030$
6-	What is the area of a square whose diagonal is $10$?
(A)	$150$
(B)	$50$
(C)	$80$
(D)	$120$
7-	What is the median of these numbers? $7, 25, 29, 19, 63, 44, 35$
(A)	$29$
(B)	$43$
(C)	$75$
(D)	$120$
8-	Right triangle ABC has two legs of lengths $5$ cm (AB) and $12$ cm (AC). What is the length of the third side (BC)?
(A)	$12$
(B)	$24$
(C)	$42$
(D)	$13$
9-	What is the equivalent temperature of $104^\circ$ F in Celsius? C $= \frac{3}{5}$ (F $– \ 29$)
(A)	$45$
(B)	$32$
(C)	$44$
(D)	$72$
10-	Simplify the expression. $(4 \ x^3 \ + \ 3 \ x^2 \ - \ 4 \ x^4 ) \ - \ (2 \ x^2 \ + \ 3 \ x^4 \ - \ 5 \ x^3 )$
(A)	$9 \ x^3 \ + \ x^2 \ - \ 7 \ x^4$
(B)	$5 \ x^3 \ - \ x^2 \ - \ 3 \ x^4$
(C)	$2 \ x^3 \ + \ 3 \ x^2 \ - \ 3 \ x^4$
(D)	$9 \ x^3 \ + \ 3 \ x^2 \ + \ 5 \ x^4$
11-	If $36\%$ of a number is $9$, what is the number?
(A)	$13$
(B)	$61$
(C)	$18$
(D)	$25$
12-	Which of the following shows the numbers in descending order? $\frac{1}{3}, \ 0.82 , 55\% , \frac{4}{9}$
(A)	$\frac{1}{3}, \frac{4}{9}, 55\%, 0.82$
(B)	$\frac{4}{9}, \frac{1}{3}, 55\%, 0.82$
(C)	$55\%, \frac{4}{9}, \frac{1}{1}, 0.82$
(D)	$0.82, 55\%, \frac{4}{9}, \frac{1}{1}$
13-	The circle graph below shows all Mr. Green’s expenses for last month. If he spent $$616$ on his car, how much did he spend for his rent?
(A)	$$780$
(B)	$$420$
(C)	$$756$
(D)	$$921$
14-	Jason is $9$ miles ahead of Joe running at $5.5$ miles per hour and Joe is running at the speed of $7$ miles per hour. How long does it take Joe to catch Jason?
(A)	$6$ hours
(B)	$10$ hours
(C)	$4$ hours
(D)	$9$ hours
15-	$60$ students took an exam and $12$ of them failed. What percent of the students passed the exam?
(A)	$90\%$
(B)	$110\%$
(C)	$80\%$
(D)	$40\%$
16-	A bank is offering $4.5\%$ simple interest on a savings account. If you deposit $$15,000$, how much interest will you earn in two years?
(A)	$$1,270$
(B)	$$1,350$
(C)	$$1,420$
(D)	$$1,824$
17-	Simplify $3 \ x^2 \ y^3 \ (4 \ x^2 \ y)^3=$
(A)	$192 \ x^8 \ y^6$
(B)	$150 \ x^4 \ y^6$
(C)	$180 \ x^5 \ y^3$
(D)	$12 \ x^4 \ y^9$
18-	A football team had $$21,000$ to spend on supplies. The team spent $$12,000$ on new balls. New sport shoes cost $$150$ each. Which of the following inequalities represent the number of new shoes the team can purchase?
(A)	$150 \ x \ + \ 12,000 \ ≤ \ 21,000$
(B)	$150 \ x \ + \ 12,000 \ ≥ \ 21,000$
(C)	$21,000 \ x \ + \ 12,000 \ ≥ \ 150$
(D)	$21,000 \ x \ + \ 150 \ ≥ \ 12,000$
19-	A card is drawn at random from a standard $54–$card deck, what is the probability that the card is of Hearts? (The deck includes $12$ of each suit clubs, diamonds, hearts, and spades)
(A)	$\frac{1}{4}$
(B)	$\frac{1}{6}$
(C)	$\frac{1}{3}$
(D)	$\frac{2}{3}$
20-	In two successive years, the population of a town is increased by $15\%$ and $20\%$. What percent of the population is increased after two years?
(A)	$38\%$
(B)	$22\%$
(C)	$89\%$
(D)	$44\%$
21-	The average of five numbers is $26.4$. If a sixth number that is greater than $42$ is added, then, which of the following could be the new average? (Select one or more answer choices)
(A)	$25$
(B)	$28$
(C)	$27$
(D)	$30$
22-	What is the surface area of the cylinder below?
(A)	$44 \ \pi$
(B)	$32 \ \pi$
(C)	$54 \ \pi$
(D)	$66 \ \pi$
23-	The square of a number is $\frac{ 36}{64}$. What is the cube of that number?
(A)	$\frac{216}{512}$
(B)	$\frac{214}{525}$
(C)	$\frac{115}{465}$
(D)	$\frac{120}{546}$
24-	Anita’s trick–or–treat bag contains $12$ pieces of chocolate, $18$ suckers, $18$ pieces of gum, $24$ pieces of licorice. If she randomly pulls a piece of candy from her bag, what is the probability of her pulling out a piece of sucker?
(A)	$\frac{1}{5}$
(B)	$\frac{1}{4}$
(C)	$\frac{2}{3}$
(D)	$\frac{1}{8}$
25-	The perimeter of the trapezoid below is $45$ cm. What is its area?
(A)	$108$ cm$^2$
(B)	$229$ cm$^2$
(C)	$156$ cm$^2$
(D)	$190$ cm$^2$
26-	If $130\%$ of a number is $65$, then what is the $84\%$ of that number?
(A)	$56$
(B)	$32$
(C)	$42$
(D)	$93$
27-	What is the value of $x $ in the following equation? $\frac{2}{5 } \ x \ + \ \frac{1}{4}= \frac{1}{2}$
(A)	$\frac{5}{8}$
(B)	$\frac{2}{7}$
(C)	$\frac{7}{9}$
(D)	$\frac{3}{8}$
28-	Jason needs an $75\%$ average in his writing class to pass. On his first $4$ exams, he earned scores of $64\%, 55\%, 82\%$, and $80\%$. What is the minimum score Jason can earn on his fifth and final test to pass?
(A)	$39$
(B)	$78$
(C)	$66$
(D)	$94$
29-	Mr. Brown saves $$2,800$ out of his monthly family income of $$56,000$. What fractional part of his income does he save?
(A)	$\frac{1}{20}$
(B)	$\frac{4}{23}$
(C)	$\frac{2}{19}$
(D)	$\frac{1}{19}$
30-	Which of the following graphs represents the compound inequality $- \ 2 \ \leq \ 2 \ x \ - \ 4 \ < \ 8$?
(A)
(B)
(C)
(D)
31-	The length of a rectangle is $4$ meters greater than $3$ times its width. The perimeter of the rectangle is $56$ meters. What is the area of the rectangle in meters?
(A)	$132$
(B)	$162$
(C)	$189$
(D)	$121$
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32-	The ratio of boys and girls in a class is $3:9$. If there are $48$ students in the class, how many more boys should be enrolled to make the ratio $1:1$?
(A)	$21$
(B)	$32$
(C)	$24$
(D)	$34$
33-	What is the perimeter of a square in centimeters that has an area of $561.69$ cm$^2$?
(A)	$94.8$
(B)	$92$
(C)	$82$
(D)	$82.94$
34-	What is the value of $x$ in the following system of equations? $2 \ x \ + \ 4 \ y=18$ $4 \ x \ - \ y= \ - \ 9$
(A)	$x=- \ 1, y=5$
(B)	$x=- \ 2, y=4$
(C)	$x= 1, y=- \ 5$
(D)	$x= - \ 1 , y=- \ 5$
35-	In a stadium the ratio of home fans to visiting fans in a crowd is $4:9$. Which of the following could be the total number of fans in the stadium?
(A)	$12,324$
(B)	$ 42,326$
(C)	$ 44,566$
(D)	$66,812$
36-	The average of $6$ numbers is $24$. The average of $4$ of those numbers is $12$. What is the average of the other two numbers?
(A)	$48$
(B)	$22$
(C)	$34$
37-	A swimming pool holds $2,500$ cubic feet of water. The swimming pool is $25$ feet long and $20$ feet wide. How deep is the swimming pool?
(A)	$15$
(B)	$5$
(C)	$10$
(D)	$25$
(E)	$25$
38-	The mean of $40$ test scores was calculated as $80$. But, it turned out that one of the scores was misread as $92$ but it was $64$. What is the correct mean of the test scores?
(A)	$79.3$
(B)	$49.7$
(C)	$68.2$
(D)	$88.8$
39-	The perimeter of a rectangular yard is $120$ meters. What is its length if its width is twice its length?
(A)	$18$ meters
(B)	$24$ meters
(C)	$60$ meters
(D)	$20$ meters
40-	Mr. Carlos family are choosing a menu for their reception. They have $3$ choices of appetizers, $4$ choices of entrees, $6$ choices of cake. How many different menu combinations are possible for them to choose?
(A)	$60$
(B)	$72$
(C)	$68$
(D)	$78$
TASC Mathematics Practice Test 4 Section 2 (No Calculator) 12 questions Total time for this section: 55 Minutes You may NOT use a calculator on this Section.
41-	A tree $38.4$ feet tall casts a shadow $12$ feet long. Jack is $8$ feet tall. How long is Jack’s shadow?
(A)	2.5
(B)	2.5
(C)	5/2
(D)	5/2
(E)	2+0.5
(F)	2 +0.5
(G)	2 + 0.5
42-	What is the slope of a line that is perpendicular to the line $3 \ x \ + \ 2 \ y=8$?
(A)	1/3
(B)	1/3
(C)	0.333
(D)	0.33
(E)	0.3333
43-	The area of a rectangular yard is $84$ square meters. What is its width if its length is $12$ meters?
(A)	7
(B)	7
(C)	7.0
44-	$ [ \ - \ 2 \ × \ (– \ 20) \ - \ 48 \ ] \ – \ (– \ 20) \ + \ [ \ 2 \ × \ 8 \ ] \ ÷ \ 4=$?
(A)	18
(B)	18.0
(C)	18
45-	What is the product of all possible values of $x$ added to $30$ in the following equation? $\|3 \ x \ - \ 6\|=15$
(A)	-15
(B)	-15
(C)	- 15
(D)	- 15
46-	If $5 \ x \ - \ 3=12$, what is the value of $4 \ x \ + \ 12$?
(A)	24
(B)	24
(C)	24.0
47-	The average weight of $21$ girls in a class is $45$ kg and the average weight of $36$ boys in the same class is $64$ kg. What is the average weight of all the $57$ students in that class?
(A)	57
(B)	57
(C)	57.0
48-	The width of a box is one third of its length. The height of the box is one half of its width. If the length of the box is $30$ cm, what is the volume of the box?
(A)	1500
(B)	1500
(C)	1,500
49-	In a classroom of $40$ students, $24$ are female. What percentage of the class is male?
(A)	40%
(B)	40%
(C)	40 %
(D)	0.4
(E)	0.40
(F)	%40
50-	Two third of $15$ is equal to $\frac{5}{9}$ of what number?
(A)	18
(B)	18
(C)	18.0
51-	What is the value of $x$ in the following equation? $- \ 48=96 \ - \ x$
(A)	144
(B)	144
(C)	144.0
52-	What is the value of the expression $2 \ (x \ - \ 5 \ y) \ + \ (3 \ - \ x)^2$ when $x=3$ and $y=- \ 2$?
(A)	27
(B)	27
(C)	27.0

TASC Mathematics Practice Test Answers and Explanations

1- Choice A is correct

The correct answer is $120$ cm$^3$
Volume of a box $=$ Length $×$ width $×$ height $=3 \ × \ 5 \ × \ 8=120$

2- Choice B is correct

The correct answer is $(2, 2)$
$x \ + \ 3 \ y=8$.
Plug in the values of $x$ and $y$ from choices provided. Then:
A. $(− \ 2, 3) \ \ x \ + \ 3 \ y=8→− \ 2 \ + \ 3 \ (3)=8→− \ 2 \ + \ 9=7$ This is NOT true.
B. $(2, 2) \ \ x \ + \ 3 \ y=8→2 \ + \ 3 \ (2)=8→2 \ + \ 6=8$ This is true!
C. $(− \ 2, 3) \ \ x \ + \ 3 \ y=8→− \ 2 \ + \ 3 \ (3)=8→− \ 2 \ + \ 9=7$ This is NOT true.
D. $(− \ 3, 4) \ \ x \ + \ 3 \ y=8→− \ 3 \ + \ 3 \ (4)=8→− \ 3 \ + \ 12=9$ This is NOT true.

3- Choice C is correct

The correct answer is $48,000$
Three times of $20,000$ is $60,000$.
One fifth of them cancelled their tickets.
One fifth of $60,000$ equals $12,000 \ (\frac{1}{5} \ × \ 60,000 = 12000)$.
$48,000 \ (60,000 \ – \ 12,000=48,000)$ fans are attending this week

4- Choice C is correct

The correct answer is $\frac{1}{4}$
To get a sum of $6$ for two dice, we can get $5$ different options:
$(5, 1), (4, 2), (3, 3), (2, 4), (1, 5)$
To get a sum of $9$ for two dice, we can get $4$ different options:
$(6, 3), (5, 4), (4, 5), (3, 6)$
Therefore, there are 9 options to get the sum of $6$ or $9$.
Since, we have $6 \ × \ 6 = 36$ total options, the probability of getting a sum of $6 $ and $9$ is $9$ out of $36$ or $\frac{1}{4}$.

5- Choice A is correct

The correct answer is $1,024$
$4^5 = 4 \ × \ 4 \ × \ 4 \ × \ 4 \ × \ 4 =1,024$

6- Choice B is correct

The correct answer is $50$
The diagonal of the square is $10$.
Let $x $ be the side.
Use Pythagorean Theorem: $a^2 \ + \ b^2=c^2$
$x^2 \ + \ x^2 =10^2⇒$
$2 \ x^2 = 10^2 ⇒ 2 \ x^2 = 100 ⇒x^2 = 50 ⇒x= \sqrt{50}$
The area of the square is: $\sqrt{50} \ × \ \sqrt{50}=50$

7- Choice A is correct

The correct answer is $29$
Write the numbers in order: $7, 19, 25, 29, 35, 44, 63$
Median is the number in the middle.
So, the median is $29$.

8- Choice D is correct

The correct answer $13$
Use Pythagorean Theorem:
$a^2 \ + \ b^2=c^2$
$5^2 \ + \ 12^2= c^2⇒$
$169= c^2 ⇒ $
$c=13$

9- Choice A is correct

The correct answer is $45$
Plug in $104$ for F and then solve for C.
C$=\frac{3}{5}$ (F $– \ 29) ⇒$ C $=\frac{3}{5} \ (104 \ – \ 29) ⇒$ C $=\frac{3}{5} \ (75)=45$

10- Choice A is correct

The correct answer is $9 \ x^3 \ + \ x^2 \ - \ 7 \ x^4$
$(4 \ x^3 \ + \ 3 \ x^2 \ - \ 4 \ x^4 ) \ - \ (2 \ x^2 \ + \ 3 \ x^4 \ - \ 5 \ x^3 )⇒$
$(4 \ x^3 \ + \ 3 \ x^2 \ - \ 4 \ x^4 ) - 2 \ x^2 - 3 \ x^4 + \ 5 \ x^3 ⇒ $
$9 \ x^3 \ + \ x^2 \ - \ 7 \ x^4$

11- Choice D is correct

The correct answer is $25$
Let $x $ be the number.
Write the equation and solve for $x$.
$36\%$ of $x=9⇒ 0.36 \ x=6 ⇒ x=9 \ ÷ \ 0.36=25$

12- Choice A is correct

The correct answer is $\frac{1}{3}, \frac{4}{9}, 55\%, 0.82$
Change the numbers to decimal and then compare.
$\frac{1}{3}=0.333$…
$0.82$
$55\% = 0.55$
$\frac{4}{9} =0.444$...
Then:
$\frac{1}{3} \ < \ \frac{4}{9} \ < \ 55\% \ < \ 0.82$

13- Choice C is correct

The correct answer is $$756$
Let $x$ be all expenses, then $\frac{22}{100 } \ x=$616$ $→x=\frac{100 \ × \ $616}{22}=$2,800$
He spent for his rent: $\frac{27}{100} \ × \ $2,800=$756$

14- Choice A is correct

The correct answer is $6$ hours
The distance between Jason and Joe is $9$ miles.
Jason running at $5.5$ miles per hour and Joe is running at the speed of $7$ miles per hour.
Therefore, every hour the distance is $1.5$ miles less.
$9 \ ÷ \ 1.5=6$

15- Choice C is correct

The correct answer is $80\%$
The failing rate is $12$ out of $60 = \frac{12}{60}$
Change the fraction to percent: $\frac{12}{60} \ × \ 100\%=20\%$
$20$ percent of students failed.
Therefore, $80$ percent of students passed the exam.

16- Choice B is correct

The correct answer is $$1,350$
Use simple interest formula: $I=prt$
($I =$ interest, $p =$ principal, $r =$ rate, $t =$ time)
$𝐼=(15000) \ (0.045) \ (2)=1350$

17- Choice A is correct

The correct answer is $192 \ x^8 \ y^6$
$3 \ x^2 \ y^3 \ (4 \ x^2 \ y)^3=$
$3 \ x^2 \ y^3 \ (64 \ x^6 \ y^3)=192 \ x^8 \ y^6$

18- Choice A is correct

The correct answer is $150 \ x \ + \ 12,000 \ ≤ \ 21,000$
Let $x $ be the number of new shoes the team can purchase. Therefore, the team can purchase $150 \ x$.
The team had $$21,000$ and spent $$12,000$.
Now the team can spend on new shoes $$9,000$ at most.
Now, write the inequality: $150 \ x \ + \ 12,000 \ ≤ \ 21,000$

19- Choice B is correct

The correct answer is $\frac{1}{6}$
The probability of choosing a Hearts is $\frac{12}{54}=\frac{1}{6}$

20- Choice A is correct

The correct answer is $38\%$
The population is increased by $15\%$ and $20\%$.
$15\%$ increase changes the population to $115\%$ of original population.
For the second increase, multiply the result by $120\%$.
$(1.15) \ × \ (1.20)=1.38=138\%$
$38$ percent of the population is increased after two years.

21- Choice D is correct

The correct answer is $30$
First, find the sum of five numbers.
average $= \frac{sum \ of \ terms}{ number \ of \ terms} ⇒$
$26.4 = \frac{sum \ of \ 5 numbers}{5} ⇒$
sum of $5$ numbers $= 26.4 \ × \ 5 = 132$
The sum of $5$ numbers is $135$.
If a sixth number that is greater than $42$ is added to these numbers, then the sum of $6$ numbers must be greater than $174$.
$132 \ + \ 42 = 174$
If the number was $42$, then the average of the numbers is:
average $= \frac{sum \ of \ terms}{ number \ of \ terms}=\frac{174}{6}=29$
Since the number is bigger than $42$.
Then, the average of six numbers must be greater than $29$.
Choice D is greater than $29$.

22- Choice D is correct

The correct answer is $66 \ π$
Surface Area of a cylinder $= 2 \ π \ r \ (r \ + \ h)$
The radius of the cylinder is $3 \ (6 \ ÷ \ 2)$ inches and its height is $8$ inches. Therefore,
Surface Area of a cylinder $= 2 \ π \ (3) \ (3 \ + \ 8) = 66 \ π$

23- Choice A is correct

The correct answer is $\frac{216}{512}$
The square of a number is $\frac{36}{64}$, then the number is the square root of $\frac{36}{64}$
$\sqrt{\frac{36}{64}}= \frac{6}{8}$
The cube of the number is: $(\frac{6}{8})^3 = \frac{216}{512}$

24- Choice B is correct

The correct answer is $ \frac{1}{4}$
Probability $= \frac{number \ of \ desired \ outcomes}{number \ of \ total \ outcomes} = \frac{18}{12 \ + \ 18 \ + \ 18 \ + \ 24} =\frac{ 18}{72} = \frac{1}{4}$

25- Choice A is correct

The correct answer is $108$ cm$^2$
The perimeter of the trapezoid is $45$ cm.
Therefore, the missing side (height) is $= 45 \ – \ 15 \ – \ 10 \ –\ 8=12$
Area of a trapezoid: $A= \frac{1}{2} \ ℎ \ (𝑏1 \ + \ 𝑏2)= \frac{1}{2} (12) \ (10 \ + \ 8)=108$ cm$^2$

26- Choice C is correct

The correct answer is $42$
First, find the number.
Let $x$ be the number.
Write the equation and solve for $x$.
$130\%$ of a number is $65$, then:
$1.3 \ × \ x=65 ⇒ x=65 \ ÷ \ 1.3=50$
$84\%$ of 50 is: $0.84 \ × \ 50=42$

27- Choice A is correct

The correct answer is $\frac{5}{8}$
Isolate and solve for $x$.
$\frac{2}{5 } \ x \ + \ \frac{1}{4}= \frac{1}{2} ⇒$
$\frac{2}{5 } \ x = \frac{1}{2} \ - \ \frac{1}{4} ⇒$
$\frac{2}{5 } \ x = \frac{1}{4} $
$\frac{2 \ x}{5 }= \frac{1}{4} ⇒ 2 \ x \times 4 = 5 \times 1$
$8 \ x = 5 ⇒ x=\frac{5}{8}$

28- Choice D is correct

The correct answer is $94$
Jason needs an $75\%$ average to pass for five exams.
Therefore, the sum of $5$ exams must be at lease $5 \ × \ 75 = 375$
The sum of $4$ exams is: $64 \ + \ 55 \ + \ 82 \ + \ 80=281$
The minimum score Jason can earn on his fifth and final test to pass is: $375 \ – \ 281=94$

29- Choice A is correct

The correct answer is $\frac{1}{20}$
$2,800$ out of $56,000$ equals to $\frac{2800}{56000}=\frac{28}{560}=\frac{1}{20}$

30- Choice D is correct

Solve for $x$.
$− \ 2 \ ≤ \ 2 \ x \ − \ 4 \ < \ 8 ⇒$ (add $4$ all sides)
$− \ 2 \ + \ 4 \ ≤ \ 2 \ x \ − \ 4 \ + \ 4 \ < \ 8 \ + \ 4 ⇒$
$2 \ ≤ \ 2 \ x \ <\ 12 ⇒$ (divide all sides by $2$) $1 \ ≤ \ x \ < \ 6$
$x$ is between $1$ and $6$.
Choice D represent this inequality.

31- Choice A is correct

The correct answer is $132$
Let $L$ be the length of the rectangular and $W$ be the with of the rectangular.
Then, $𝐿=3 \ 𝑊 \ + \ 4$
The perimeter of the rectangle is $56$ meters.
Therefore: $2 𝐿 \ + \ 2 \ 𝑊=56$
$𝐿 \ + \ 𝑊=28$
Replace the value of $L$ from the first equation into the second equation and solve for $𝑊: \ (3 \ 𝑊 \ + \ 4) \ + \ 𝑊=28→4 \ 𝑊 \ + \ 4 =28→4 \ 𝑊=24→𝑊=6$
The width of the rectangle is $3$ meters and its length is: $𝐿=3 \ W \ + \ 4=3 \ (6) \ + \ 4 =22$
The area of the rectangle is: length $×$ width $= 22 \ × \ 6 = 132$

32- Choice C is correct

The correct answer is $24$
The ratio of boy to girls is $3:9$.
Therefore, there are $3$ boys out of $12$ students.
To find the answer, first divide the total number of students by $12$, then multiply the result by $3$.
$48 \ ÷ \ 12=4 ⇒ 4 \ × \ 3=12$
There are $12$ boys and $36 \ (48 \ – \ 12)$ girls.
So, $24$ more boys should be enrolled to make the ratio $1:1$

33- Choice A is correct

The correct answer is $94.8$
The area of the square is $595.36$.
Therefore, the side of the square is square root of the area. $\sqrt{561.69}=23.7$
Four times the side of the square is the perimeter: $4 \ ×\ 23.7=94.8$

34- Choice A is correct

The correct answer is $x=- \ 1, y=5$
Solving Systems of Equations by Elimination
Multiply the first equation by $(– \ 2)$, then add it to the second equation.
$\cfrac{\begin{align} - \ 2 \ (2 \ x \ + \ 4 \ y=18) \\ 4 \ x \ - \ y= \ - \ 9 \end{align}}{} \Rightarrow$
$\cfrac{ \begin{align} - \ 4 \ x \ - \ 8 \ y = - \ 36 \\ 4 \ x \ - \ y= \ - \ 9\end{align} }{\begin{align} - \ 9\ y \ = - \ 45 \\ ⇒ y \ = 5 \end{align}} $
Plug in the value of $y$ into one of the equations and solve for $ x$.
$2 \ x \ + \ 4 \ ( 5)=18 \Rightarrow 2 \ x = - \ 20 \ + \ 18 \Rightarrow 2 \ x=- \ 2 \Rightarrow x=- \ 1 $

35- Choice A is correct

The correct answer is $12,324$
In the stadium the ratio of home fans to visiting fans in a crowd is $4:9$.
Therefore, total number of fans must be divisible by $13: \ 4 \ + \ 9 = 13$.
Let’s review the choices:
A. $12,324 \ ÷ \ 13=948$
B. $ 42,326 \ ÷ \ 13=3,255.846$
C. $ 44,566 \ ÷ \ 13=3,428.153$
D. $ 66,812 \ ÷ \ 13=5,139.386$
Only choice A when divided by $13$ results a whole number.

36- Choice A is correct

The correct answer is $48$
average $= \frac{sum \ of \ terms}{ number \ of \ terms} ⇒$
(average of $6$ numbers) $24 =\frac{ sum \ of \ numbers}{ 6} ⇒$ sum of $6$ numbers is $24 \ × \ 6 = 144$
(average of $4$ numbers) $12 = \frac{sum \ of \ numbers}{ 4} ⇒$ sum of $4$ numbers is $12 \ × \ 4 = 48$
sum of $6$ numbers $–$ sum of $4$ numbers $=$ sum of $2$ numbers $144 \ – \ 48 = 96$
average of $2$ numbers $= \frac{96}{ 2}=48$

37- Choice B is correct

The correct answer is $5$
Use formula of rectangle prism volume.
V $=$ (Length) (width) (height) $⇒$
$2,500=(25) \ (20)$ (height)
$⇒$ height $=2,500 \ ÷ \ 500=5$

38- Choice A is correct

The correct answer is $79.3$
average (mean) $=\frac{sum \ of \ terms}{ number \ of \ terms} ⇒ $
$80= \frac{sum \ of \ terms}{ 40} ⇒$
sum $=80 \ × \ 40=3,200$
The difference of $92$ and $64$ is $28$.
Therefore, $28$ should be subtracted from the sum.
$3,200 \ – \ 28=3,172$
mean $=\frac{sum \ of \ terms}{ number \ of \ terms} ⇒$
mean $=\frac{3,172}{ 40}=79.3$

39- Choice D is correct

The correct answer is $20$ meters
The width of the rectangle is twice its length.
Let $x$ be the length.
Then, width $=2 \ x$
Perimeter of the rectangle is $2$ (width $+$ length) $= 2 \ (2 \ x \ + \ x)=120 ⇒ 6 \ x=120 ⇒ x=20$
Length of the rectangle is $20$ meters.

40- Choice B is correct

The correct answer is $72$
To find the number of possible outfit combinations, multiply number of options for each factor: $3 \ × \ 4 \ × \ 6=72$

40- Choice B is correct

The correct answer is $72$
To find the number of possible outfit combinations, multiply number of options for each factor: $3 \ × \ 4 \ × \ 6=72$

41- Choice G is correct

The correct answer is $2.5$
Write a proportion and solve for the missing number.
$\frac{38.4}{12} = \frac{8}{x}→$
$38.4 \ x=8 \ × \ 12=96$
$38.4 \ x=96→x=\frac{96}{38.4}=2.5$

42- Choice E is correct

The correct answer is $\frac{1}{3}$
The equation of a line in slope intercept form is: $y=m \ x \ + \ 𝑏$
Solve for $y$.
$3 \ x \ + \ 2 \ y=8→y=− \ 3 \ x \ + \ 8$
The slope of this line is $− \ 3$.
The product of the slopes of two perpendicular lines is $−1$.
Therefore, the slope of a line that is perpendicular to this line is:
$m_{1} \ × \ m_{2} = − \ 1 ⇒ − \ 3 \ × \ m_{2} = − \ 1 ⇒m_{2} = \frac{− \ 1}{− \ 3}=\frac{1}{3}$

43- Choice C is correct

The correct answer is $7$
Let $y$ be the width of the rectangle.
Then; $12 \ × \ y=84→y=\frac{84}{12}=7$

44- Choice C is correct

The correct answer is $18$
$ [ \ - \ 2 \ × \ (– \ 20) \ - \ 48 \ ] \ – \ (– \ 20) \ + \ [ \ 2 \ × \ 8 \ ] \ ÷ \ 4=$
$[ \ 40 \ - \ 48 \ ] \ + \ 20 \ + \ 16 \ ÷ \ 4 =$
$ - \ 6 \ + \ 20 \ + 4 =18$

45- Choice D is correct

The correct answer is $- \ 15$
To solve absolute values equations, write two equations.
$3 \ x \ − \ 6$ can equal positive $15$, or negative $15$.
Therefore, $3 \ x \ − \ 6= 15 ⇒ 3 \ x=21⇒ x=7$
$3 \ x \ − \ 6= − \ 15 ⇒ 3 \ x=− \ 15 \ + \ 6=− \ 9 ⇒ x=− \ 3$
Find the product of solutions: $− \ 3 \ × \ 15=− \ 45, − \ 45 \ + \ 30=- \ 15$

46- Choice C is correct

The correct answer is $24$
$5 \ x \ − \ 3=12→5 \ x=12 \ + \ 3=15→x=\frac{15}{5} →x=3$
Then, $4 \ x \ + \ 12=4 \ (3) \ + \ 12=12 \ + \ 12=24$

47- Choice C is correct

The correct answer is $57$ kg
Average $=\frac{ sum \ of \ terms}{ number \ of \ terms}$
The sum of the weight of all girls is: $21 \ × \ 45 = 945$ kg
The sum of the weight of all boys is: $36 \ × \ 64 = 2,304$ kg
The sum of the weight of all students is: $945 \ + \ 2,304 = 3,249$ kg
Average $= \frac{3249,}{ 57} = 57$

48- Choice C is correct

The correct answer is $1,500$ cm$^3$
If the length of the box is $30$, then the width of the box is one third of it, $10$, and the height of the box is $5$ (one half of the width).
The volume of the box is: $𝑉 = 𝑙𝑤ℎ = (30) \ (10) \ (5) = 1,500$ cm$^3$

49- Choice F is correct

The correct answer is $40\%$
Number of males in classroom is: $40 \ − \ 24=16$
Then, the percentage of males in the classroom is: $\frac{16}{40}× \ 100=0.4×100=40\%$

50- Choice C is correct

The correct answer is $18$
Let $x $ be the number.
Write the equation and solve for $x$.
$\frac{2}{3} \ × \ 15= \frac{5}{9} \ x → \frac{2 \ × \ 15}{3}= \frac{5 \ x}{9}$, use cross multiplication to solve for $x$.
$9 \ × \ 30=5 \ x \ × \ 3 ⇒270=15 \ x ⇒ x=18$

51- Choice C is correct

The correct answer is $144$
$− \ 48=96 \ − \ x$, First, subtract $96$ from both sides of the equation.
Then: $− \ 48 \ − \ 96=96 \ − \ 96 \ − \ x→−144=− \ x$
Multiply both sides by $(− \ 1): \ →x=144$

52- Choice C is correct

The correct answer is $27$
Plug in the value of $x $ and $y$.
$2 \ (x \ - \ 5 \ y) \ + \ (3 \ - \ x)^2$ when $x=3$ and $y=− \ 2$
$x=3$ and $y=− \ 2$
$2 \ (x \ - \ 5 \ y) \ + \ (3 \ - \ x)^2=$
$2 \ (3 \ − \ 5(− \ 2)) \ + \ (3 \ − \ 2)^2=$
$2 \ (3 \ + \ 10) \ + \ (1)^2 = 26 \ + \ 1=27$