10 Most Common 8th Grade ACT Aspire Math Questions

B. 40

C. 64

D. 80

3- A $40 shirt now selling for $28 is discounted by what percent?

A. \(20 \%\)

B. \(30 \%\)

C. \(40 \%\)

D. \(60 \%\)

4- How much interest is earned on a principal of $5000 invested at an interest rate of \(5\%\) for four years?

A. $250

B. $500

C. $1000

D. $2000

5- A swimming pool holds 2,000 cubic feet of water. The swimming pool is 25 feet long and 10 feet wide. How deep is the swimming pool?
Write your answer in the box below. ________

6- The price of a car was $20,000 in 2014, $16,000 in 2015 and $12,800 in 2016. What is the rate of depreciation of the price of car per year?

A. \(15 \%\)

B. \(20 \%\)

C. \(25 \%\)

D. \(30 \%\)

7- What is the area of the shaded region if the diameter of the bigger circle is 12 inches and the diameter of the smaller circle is 8 inches.

A. \(16 π\)

B. \(20 π\)

C. \(36 π\)

D. \(80 π\)

8- What is the area of an isosceles right triangle that has one leg that measures 6 cm? _________

9- A taxi driver earns $9 per 1-hour work. If he works 10 hours a day and in 1 hour he uses 2-liters petrol with price $1 for 1-liter. How much money does he earn in one day?

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A. $90

B. $88

C. $70

D. $60

10- What is the solution of the following system of equations?
\(\frac{-x}{2}+ \frac{y}{4} = 1\)
\(\frac{-5y}{6}+ 2x = 4\)

A. \(x=48,y=22 \)

B. \(x=50,y=20\)

C. \(x=20,y=50\)

D. \(x=22,y=48\)

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Answers:

2- B
Use the area of rectangle formula \((s = a × b)\).
To find the area of the shaded region subtracts the smaller rectangle from the bigger rectangle.
\(S_{1} – S_{2} = (10 ft × 8ft) – (5ft × 8ft) ⇒ S_{1} – S_{2} = 40ft\)

3- B
Use the formula for Percent of Change
\(\frac{New \space Value-Old \space Value}{Old \space Value}× 100 \%\)
\(\frac{28-40}{40}× 100 \% = –30 \% \)(negative sign here means that the new price is less than old price).

4- C
Use simple interest formula:
I=prt
(I = interest, p = principal, r = rate, t = time)
\(I=(5000)(0.05)(4)=1000\)

5- 8
Use formula of rectangle prism volume.
\(V = (length) (width) (height) ⇒ 2000 = (25) (10) (height) ⇒
height = 2000 ÷ 250 = 8\)

6- B
Use this formula: Percent of Change
\(\frac{New \space Value-Old \space Value}{Old \space Value}× 100 %\)
\(\frac{16000-20000}{20000}× 100 % \)
\(= 20 \% \space and \space \frac{12800-16000}{16000}× 100 \% = 20 \%\)

7- B
To find the area of the shaded region subtract smaller circle from bigger circle.
\(S_{bigger}-S_{smaller} =π(r_{bigger})^2 -π(r_{smaller})^2⇒S_{bigger}-S_{smaller}=π(6)^2-π(4)^2\)
\(⇒ 36 π – 16π = 20 π\)

8- 18
\(a = 6 ⇒\) area of the triangle is:
\(\frac{1}{2}(6×6)=\frac{36}{2}=18 \space cm^2\)

9- C
\($9×10=$90\)
Petrol use: \(10×2=20 \space liters\)
Petrol cost: \(20×$1=$20\)
Money earned: \($90-$20=$70\)

10- D
\(\frac{-x}{2}+ \frac{y}{4} = 1\)
\(\frac{-5y}{6}+ 2x = 4\)
Multiply the top equation by 4. Then,
\(-2x+ y = 4\)
\(\frac{-5y}{6}+ 2x = 4\)
Add two equations.
\(\frac{1}{6}y=8→y=48\), plug in the value of y into the first equation →\(x=22\)

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