# HSPT Math Practice Test Questions

Preparing for the HSPT Math test? Try these free HSPT Math Practice questions. Reviewing practice questions is the best way to brush up on your Math skills. Here, we walk you through solving 10 common HSPT Math practice problems covering the most important math concepts on the HSPT Math test.

These HSPT Math practice questions are designed to be similar to those found on the real HSPT Math test. They will assess your level of preparation and will give you a better idea of what to study for your exam.

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**10 Sample HSPT Math Practice Questions**

1- Which of the following points lies on the line \(3x+2y=11\)?

A. \((-1,3)\)

B. \( (2,3)\)

C. \((-1,7)\)

D. \((0,2)\)

2- Two-thirds of \(9\) is equal to \(\frac{2}{5}\) of what number?

A. 5

B. 9

C. 15

D. 25

3- The marked price of a computer is D dollar. Its price decreased by \(15\%\) 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. 0.93 D

C. 0.97 D

D. 1.20 D

4- A $45 shirt now selling for $28 is discounted by what percent?

A. \(20\%\)

B. \(37.7\%\)

C. \(40.5\%\)

D. \(60\%\)

5- Which of the following could be the product of two consecutive prime numbers?

A. 2

B. 10

C. 24

D. 35

6- Which of the following lists shows the fractions in order from least to greatest?

\(\frac{5}{7},\frac{1}{7},\frac{3}{8},\frac{5}{11}\)

A. \(\frac{3}{8},\frac{1}{7},\frac{5}{7},\frac{5}{11}\)

B. \(\frac{1}{7},\frac{5}{11},\frac{3}{8},\frac{5}{7}\)

C. \(\frac{1}{7},\frac{3}{8},\frac{5}{11},\frac{5}{7}\)

D. \(\frac{3}{8},\frac{1}{7},\frac{5}{11},\frac{5}{7}\)

7- A boat sails \(120\) miles south and then \(50\) miles east. How far is the boat from its start point?

A. \(45\) miles

B. \(130\) miles

C. \(160\) miles

D. \(170\) miles

8- The ratio of boys and girls in a class is \(4:7\). If there are \(55\) students in the class, how many more boys should be enrolled to make the ratio \(1:1\)?

A. 8

B. 10

C. 15

D. 20

9- Sophia purchased a sofa for \($530.20\). The sofa is regularly priced at \($631\). What was the percent discount Sophia received on the sofa?

A. \(12\%\)

B. \(16\%\)

C. \(20\%\)

D. \(25\%\)

10- The score of Emma was half as that of Ava and the score of Mia was twice that of Ava. If the score of Mia was \(40\), what is the score of Emma?

A. 5

B. 10

C. 20

D. 40

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## Answers:

1- **C**

\(3x+2y=11\). Plug in the values of \(x \) and \(y\) from choices provided. Then:

☐A. \((-1,3) \ \ \ 3x+2y=11→3(-1)+2(3)=11→-3+6=11\) NOT true!

☐B. \((2,3) \ \ \ 3x+2y=11→3(2)+2(3)=11→6+6=11\) NOT true!

☐C. \((-1,7) \ \ \ 3x+2y=11→3(-1)+2(7)=11→-3+14=11\) Bingo!

☐D. \((0,2) \ \ \ 3x+2y=11→3(0)+2(2)=11→0+4=11\) Nope!

2- **C**

Let \(x\) be the number. Write the equation and solve for \(x\).

\(\frac{2}{3}×9= \frac{2}{5 } . x ⇒ \frac{2×9}{3}= \frac{2x}{5}\) , use cross multiplication to solve for \(x\).

\(5×18=2x×3 ⇒90=6x ⇒ x=15\)

3-** B**

To find the discount, multiply the number by (\(100\%-\)rate of discount).

Therefore, for the first discount we get: \((D) (100\%-15\%) = (D) (0.85) = 0.85 D\)

For increase of \(10\%: (0.85 D) (100\%+10\%)=(0.85 D) (1.10)=0.93D =93\%\) of D

4- **B**Use the formula for Percent of Change \(\frac{New \ Value-Old \ Value}{Old \ Value}×100\% \)

\(\frac{28-45}{45}×100\%=-37.7\%\) (Negative sign here means that the new price is less than the old price).

5- **D**

Some of prime numbers are: \(2,3,5,7,11,13\). Find the product of two consecutive prime numbers: \(2×3=6\) (not in the options),\(3×5=15\) (not in the options),\(5×7=35\) (bingo!),

\(7×11=77\) (not in the options)

6- **C**

Let’s compare each fraction:\(\frac{1}{7} < \frac{3}{8} < \frac{5}{11} < \frac{5}{7}\). Only choice C provides the right order.

7- **B**

Use the information provided in the question to draw the shape.

Use Pythagorean Theorem: \(a^2+ b^2=c^2 \)

\(120^2+50^2=c^2 ⇒ 14400+2500= c^2 ⇒ 16900=c^2 ⇒ c=130 \)

8- **C**

The ratio of boys to girls is \(4:7\). Therefore, there are \(4\) boys out of \(11\) students. To find the answer, first, divide the total number of students by \(11\), then multiply the result by \(4. 55÷11=5 ⇒ 4×5=20\). There are \(20\) boys and \(35 (55-20) \)girls. So, \(15\) more boys should be enrolled to make the ratio \(1:1\)

9- **B**

The question is this: \(530.20\) is what percent of \(631\)? Use percent formula:

part\(=\frac{percent}{100}×\)whole. \(530.20=\frac{percent}{100}×631 ⇒ 530.20= \frac{percent ×631}{100} ⇒ 53020 =\)percent \(×631 ⇒\) percent\(=\frac{53020}{631}=84.02≅84\) .

\(530.20\) is \(84\%\) of \(631\). Therefore, the discount is: \(100\%-84\%=16\%\)

10- **B**

If the score of Mia was \(40\), therefore the score of Ava is \(20\). Since the score of Emma was half as that of Ava, therefore, the score of Emma is \(10\).

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