FREE TASC Math Practice Test

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10 Sample TASC Math Practice Questions

1- If \(3x-5=8.5\), What is the value of \(5x+3\) ? __________

2- What is the area of an isosceles right triangle that has one leg that measures \(4\)? __________

3- The perimeter of the trapezoid below is \(50\). What is its area? __________
\(\img{https://appmanager.effortlessmath.com/public/images/questions/rt8899kkkk.png
}\)

4- From last year, the price of gasoline has increased from \($1.40\) per gallon to \($1.75\) per gallon. The new price is what percent of the original price? __________

5- \(-18+6×(-5)-[4+22×(-4)]÷2+8=\) ? __________

6- A ladder leans against a wall forming a \(60^\circ\) angle between the ground and the ladder. If the bottom of the ladder is \(30\) feet away from the wall, how many feet is the ladder? __________

7- The volume of cube A is \(\frac{1}{2}\) of its surface area. What is the length of an edge of cube A? _________

8- What is the value of \(f(5)\) for the following function \(f\)?__________
\(f(x)=x^2-3x\)

9- If \(\frac{x-3}{5}=N\) and \(N=6\), what is the value of \(x\)? __________

10- If the ratio of \(5a\) to \(2b\) is \(\frac{1}{4}\), what is the ratio of a to b? __________

11- A construction company is building a wall. The company can build \(30\) cm of the wall per minute. After \(50\) minutes \(\frac{3}{4}\) of the wall is completed. How many meters is the wall? __________

12- The average of \(13,15,20\) and \(x\) is \(20\). What is the value of \(x\)? __________

13- In the figure below, what is the value of \(x\)?
\(\img{https://appmanager.effortlessmath.com/public/images/questions/fffuuuu.png
}\)

A. 43

B. 67

C. 77

D. 90

14- Simplify.
\((8x^3-8x^2+2x^4 )-(4x^2-2x^4+2x^3 )\)

A. \(4x^4+6x^3-12x^2\)

B. \(4x^3-12x^2\)

C. \(4x^4+4x^3-12x^2\)

D. \(8x^3-12x^2\)

15- 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. \(32\%\)

B. \(35\%\)

C. \(38\%\)

D. \(68\%\)

16- Which of the following answers represents the compound inequality \(-2≤2x-4<8\)?

A. \( 2 < x < 4\)

B. \(2≤x≤4\)

C. \(1 < x ≤ 6\)

D. \(1≤x<6\)

17- What is the volume of a box with the following dimensions?
Hight = \(4\) cm Width = \(5\) cm Length = \(6\) cm

A. \(15\) cm \(^3\)

B. \(60\) cm \(^3\)

C. \(90\) cm \(^3\)

D. \(120\) cm \(^3\)

18- Mr. Carlos’s family is choosing a menu for their reception. They have \(6\) choices of appetizers, \(5\) choices of entrees, \(4\) choices of cake. How many different menu combinations are possible for them to choose from?

A. 12

B. 32

C. 120

D. 150

19- In a stadium, the ratio of home fans to visiting fans in a crowd is \(5:7\). 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

20- Last week \(24,000\) fans attended a football match. This week three times as many bought tickets, but one-sixth of them canceled their tickets. How many are attending this week?

A. 48,000

B. 54,000

C. 60,000

D. 72,000

Best TASC Math Prep Resource for 2022

Answers:

2- 8
First, draw an isosceles triangle. Remember that the two sides of the triangle are equal. Let put a for the legs. Then:
\(a=4⇒\) area of the triangle is \(=\frac{1}{2} (4×4)=\frac{16}{2}=8\)
\(\img{https://appmanager.effortlessmath.com/public/images/questions/dtawyy4wy5.png
}\)

3- 78
The perimeter of the trapezoid is \(50\).
Therefore, the missing side (height) is \(= 50-18-12-14 =6\)
Area of a trapezoid:
A \(= \frac{1}{2} h (b_{1}+b_{2}) = \frac{1}{2} (6) (12+14) =78\)

4- 125
The question is this: \(1.75\) is what percent of \(1.40\)? Use percent formula:
part\(=\frac{percent}{100}×\)whole\(⇒1.75=\frac{percent}{100}×1.40 ⇒ 1.75=\frac{percent ×1.40}{100} ⇒175=\)percent\(×1.40 ⇒\) percent \(= \frac{175}{1.40} = 125\)

5- 2
Use PEMDAS (order of operation):
\(-18+6×(-5)-[4+22×(-4)]÷2+8=-18-30-[4-88]÷2+8=-48-[-84]÷2+8=-48+84÷2+8=-48+42+8=2\)

6- 60
The relationship among all sides of a special right triangle
\(30^\circ-60^\circ-90^\circ\) is provided in this triangle:
In this triangle, the opposite side of \(30^\circ\) angle is half of the hypotenuse.
Draw the shape for this question:
The latter is the hypotenuse. Therefore, the latter is \(60\) ft.
\(\img{https://appmanager.effortlessmath.com/public/images/questions/erw.png
}\)

7- 3
Let \(x\) be the length of an edge of a cube, then the volume of a cube is: \(V=x^3 \)
The surface area of a cube is: \(SA=6x^2\)
The volume of cube A is \(\frac{1}{2}\) of its surface area. Then: \(x^3=\frac{6x^2}{2}→x^3=3x^2\), divide both side of the equation by \(x^2\). Then: \(\frac{x^3}{x^2} =\frac{3x^2}{x^2} →x=3\)

8- 10
The input value is \(5\). Then: \(x=5, f(x)=x^2-3x→ f(5)=5^2-3(5)=25-15=10\)

9- 33
Since \(N=6\), substitute \(6\) for N in the equation \(\frac{x-3}{5}=N\), which gives \(\frac{x-3}{5}=6\). Multiplying both sides of \(\frac{x-3}{5}=6\) by \(5\) gives \(x-3=30\) and then adding \(3\) to both sides of \(x-3=30\) then, \(x=33\).

10- 0.1
Write the ratio of \(5a\) to \(2b. \frac{5a}{2b}=\frac{1}{4}\). Use cross multiplication and then simplify. \(5a×4=2b×1→20a=2b→a=\frac{2b}{20}=\frac{b}{10}\). Now, find the ratio of a to \(b\). \(\frac{a}{b}=\frac{\frac{b}{10}}{b}→\frac{b}{10}÷b=\frac{b}{10}×\frac{1}{b}=\frac{b}{10b}=\frac{1}{10}=0.1\)

11- 20
The rate of construction company\(=\frac{30 \ cm}{1 \ min}=30 \ \frac{cm}{min }\)
Height of the wall after \(50\) minutes \(= \frac{30 \ cm}{1 \ min}×50\) min\(=1,500\) cm
Let \(x\) be the height of wall, then \(\frac{3}{4} x=1,500\) cm \(→x=\frac{4×1,500}{3}→x=2,000\) cm\(=20\) m

12- 32
average\(=\frac{sum \ of \ terms}{number \ of \ terms}⇒ 20=\frac{13+15+20+x}{4}⇒80=48+x⇒x=32\)

13- B
\(α=180^\circ-112^\circ=68^\circ\)
\(β=180^\circ-135^\circ=45^\circ\)
\(x+α+β=180^\circ→x=180^\circ-68^\circ-45^\circ=67^\circ\)
\(\img{https://appmanager.effortlessmath.com/public/images/questions/ertw43.png
}\)

14- A
Simplify and combine like terms.
\((8x^3-8x^2+2x^4 )-(4x^2-2x^4+2x^3 ) ⇒ (8x^3-8x^2+2x^4 )-4x^2+2x^4-2x^3⇒\)
\(4x^4+6x^3-12x^2\)

15- C
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.

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

17- D
Volume of a box\(=\)length\(×\)width\(×\)height\(=4×5×6=120\)

18- C
To find the number of possible outfit combinations, multiply the number of options for each factor: \(6×5×4=120\)

19- A
In the stadium, the ratio of home fans to visiting fans in a crowd is \(5:7\). Therefore, the total number of fans must be divisible by \(12: 5+7 =12\).
Let’s review the choices:
A. \(12,324: 12,324÷12=1,027\)
B. \(42,326 : 42,326÷12=3,527.166\)
C. \(44,566 : 44,566÷12=3,713.833\)
D. \(66,812 : 66,812÷12=5,567.666\)
Only choice A, when divided by \(12\), results in a whole number.

20- C
Three times of \(24,000\) is \(72,000\). One sixth of them cancelled their tickets. One sixth of \(72,000\) equals \(12,000 (\frac{1}{6} ×72000 =12000). 60,000(72,000-12,000=60,000) \) fans are attending this week.

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