Full Length TASC Mathematics Practice Test

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

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.

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

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 $\frac{1}{4}$.

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

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: $\sqrt{50} \ × \ \sqrt{50}=50$

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 $29$.

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

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 $=\frac{3}{5} \ (75)=45$

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

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

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:
$\frac{1}{3} \ < \ \frac{4}{9} \ < \ 55\% \ < \ 0.82$

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=$756$

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.
$9 \ ÷ \ 1.5=6$

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.

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

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

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: $150 \ x \ + \ 12,000 \ ≤ \ 21,000$

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

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.

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 $29$.

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 $= 2 \ π \ (3) \ (3 \ + \ 8) = 66 \ π$

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: $(\frac{6}{8})^3 = \frac{216}{512}$

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

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$^2$

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: $0.84 \ × \ 50=42$

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

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: $375 \ – \ 281=94$

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

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.

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 $= 22 \ × \ 6 = 132$

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 $1:1$

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: $4 \ ×\ 23.7=94.8$

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 $x$.
$2 \ x \ + \ 4 \ ( 5)=18 \Rightarrow 2 \ x = - \ 20 \ + \ 18 \Rightarrow 2 \ x=- \ 2 \Rightarrow x=- \ 1$

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.

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 $= \frac{96}{ 2}=48$

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

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 $=\frac{3,172}{ 40}=79.3$

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.

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

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

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

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:
$m_{1} \ × \ m_{2} = − \ 1 ⇒ − \ 3 \ × \ m_{2} = − \ 1 ⇒m_{2} = \frac{− \ 1}{− \ 3}=\frac{1}{3}$

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

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

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: $− \ 3 \ × \ 15=− \ 45, − \ 45 \ + \ 30=- \ 15$

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

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 $= \frac{3249,}{ 57} = 57$

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$^3$

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

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 $x$.
$9 \ × \ 30=5 \ x \ × \ 3 ⇒270=15 \ x ⇒ x=18$

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 $(− \ 1): \ →x=144$

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 $y=− \ 2$
$2 \ (x \ - \ 5 \ y) \ + \ (3 \ - \ x)^2=$
$2 \ (3 \ − \ 5(− \ 2)) \ + \ (3 \ − \ 2)^2=$
$2 \ (3 \ + \ 10) \ + \ (1)^2 = 26 \ + \ 1=27$