10 Most Common HiSET Math Questions

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10 Sample HiSET Math Practice Questions

1- The ratio of boys to girls in a school is 2:3. If there are 600 students in a school, how many boys are in the school?

A. 540

B. 360

C. 300

D. 280

E. 240

2- 25 is What percent of 20?

A. \(20 \%\)

B. \(25 \%\)

C. \(125 \%\)

D. \(150 \%\)

E. \(300 \%\)

3- The perimeter of the trapezoid below is 54. What is its area?

A. \(252 cm^2\)

B. \(234 cm^2\)

C. \(216 cm^2\)

D. \(154 cm^2\)

E. \(130 cm^2\)

4- Two-thirds of 18 is equal to \(\frac{2}{5}\) of what number?

A. 12

B. 20

C. 30

D. 60

E. 90

5- The marked price of a computer is D dollar. Its price decreased by \(20\%\) in January and later increased by \(10 \%\) in February. What is the final price of the computer in D dollars?

A. 0.80 D

B. 0.88 D

C. 0.90 D

D. 1.20 D

E. 1.40 D

6- The area of a circle is \(64 π\). What is the circumference of the circle?

A. \(8 π\)

B. \(16 π\)

C. \(32 π\)

D. \(64 π\)

E. \(124 π\)

7- A \($40\) shirt now selling for \($28\) is discounted by what percent?

A. \(20 \%\)

B. \(30 \%\)

C. \(40 \%\)

D. \(60 \%\)

E. \(80 \%\)

8- In 1999, the average worker’s income increased by $2,000 per year starting from $24,000 annual salary. Which equation represents income greater than average? (\(I =\) income, \(x =\) number of years after 1999)

A. \(I > 2000 x + 24000\)

B. \(I > – 2000 x + 24000\)

C. \(I < –2000 x + 24000\)

D. \(I < 2000 x – 24000\)

E. \(I < 24,000 x + 24000\)

9- From last year, the price of gasoline has increased from $1.25 per gallon to $1.75 per gallon. The new price is what percent of the original price?

A. \(72 \%\)

B. \(120 \%\)

C. \(140 \%\)

D. \(160 \%\)

E. \(180 \%\)

10- A boat sails 40 miles south and then 30 miles east. How far is the boat from its start point?

A. 45 miles

B. 50 miles

C. 60 miles

D. 70 miles

E. 80 miles

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Answers:

2- C
Use percent formula:
\(part = \frac{percent}{100} × whole\)
\(25 = \frac{percent}{100} × 20 ⇒ 25 = \frac{percent ×20}{100}\)
\(⇒ 25 = \frac{percent ×2}{10}\)
multiply both sides by 10.
\(4250 = percent ×2, divide \space both \space sides \space by \space 2. \)
\(125 = percent\)

3- E
The perimeter of the trapezoid is 54.
Therefore, the missing side (height) is \(= 54 – 18 – 12 – 14 = 10\)
Area of the trapezoid:
\(A = \frac{1}{2}h (b_{1} + b_{2}) = \frac{1}{2}(10)(12 + 14) = 130\)

4- C
Let \(x\) be the number. Write the equation and solve for \(x\).
\(\frac{2}{3} ×18=\frac{2}{5} ⇒\frac{2×18}{3} = \frac{2x}{5} \)
use cross multiplication to solve for \(x\).
\(5×36=2x×3 ⇒180=6x ⇒ x=30\)

5- B
To find the discount, multiply the number by \((100\% – rate of discount)\).
Therefore, for the first discount we get: (D) \((100\% – 20\%) =\) (D) \((0.80) = 0.80\) D
For increase of \(10 \%:\) (0.85 D) \((100\% + 10\%) =\) (0.85 D) \((1.10) = 0.88\) D \(= 88\%\) of D

6- B
Use the formula of areas of circles.
\(Area = πr^2 ⇒ 64 π = πr^2 ⇒ 64 = r^2 ⇒ r = 8\)
The radius of the circle is 8. Now, use the circumference formula:
Circumference \(= 2πr = 2π (8) = 16 π\)

7- 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 the old price).

8- A
Let \(x\) be the number of years. Therefore, $2,000 per year equals \(2000x\).
starting from $24,000 annual salary means you should add that amount to \(2000x\).
Income more than that is:
\(I > 2000 x + 24000\)

9- C
The question is this: 1.75 is what percent of 1.25?
Use percent formula:
\(part =\frac{percent}{100}× whole\)
\(1.75 =\frac{percent}{100}× 1.25\)
\(1.75 =\frac{percent ×1.25}{100} \)
\(175 = percent ×1.25 ⇒\)
\(percent = \frac{175}{1.25}= 140\)

10- B
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: \(a^2 + b^2 = c^2\)
\(40^2 + 30^2 = c^2 ⇒ 1600 + 900 = c^2 ⇒ 2500 = c^2 ⇒ c = 50\)

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