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