PSAT Math FREE Sample Practice Questions

PSAT Math FREE Sample Practice Questions

Preparing for the PSAT Math test? To do your best on the PSAT Math test, you need to review and practice real PSAT Math questions.  There’s nothing like working on PSAT Math sample questions to hone your math skills and put you more at ease when taking the PSAT Math test. The sample math questions you’ll find here are brief samples designed to give you the insights you need to be as prepared as possible for your PSAT Math test.

Check out our sample PSAT Math practice questions to find out what areas you need to practice more before taking the PSAT Math test!

Start preparing for the 2020 PSAT Math test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions you need to practice.

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10 Sample PSAT Math Practice Questions

1- Types of air pollutions in 10 cities of a country

What percent of cities are in the type of pollution \(A, C,\) and \(E\) respectively

☐A. \(60\%, 40\%, 90\%\)

☐B. \(30\%, 40\%, 90\%\)

☐C. \(30\%, 40\%, 60\%\)

☐D. \(40\%, 60\%, 90\%\)

2- Types of air pollutions in 10 cities of a country

How many cities should be added to type of pollutions \(B\) until the ratio of cities in type of pollution \(B\) to cities in type of pollution \(E\) will be 0.625?

☐A. 2

☐B. 3

☐C. 4

☐D. 5

3- Types of air pollutions in 10 cities of a country

If a is the mean (average) of the number of cities in each pollution type category, b is the mode, and c is the median of the number of cities in each pollution type category, then which of the following must be true?

☐A. \(a < b< c\)

☐B. \(b < a < c\)

☐C. \(a = c\)

☐D. \(b < c = a\)

4- The ratio of boys and girls in a class is \(4:7\). If there are 44 students in the class, how many more boys should be enrolled to make the ratio \(1:1\)?

☐A. 8

☐B. 10

☐C. 12

☐D. 16

5- In the following right triangle, if the sides AB and AC become twice longer, what will be the ratio of the perimeter of the triangle to its area?

☐A. \(\frac{1}{2}\)

☐B. 2

☐C. \(\frac{1}{3}\)

☐D. 3

6- The capacity of a red box is \(20\%\) bigger than the capacity of a blue box. If the red box can hold 30 equal sized books, how many of the same books can the blue box hold?

☐A. 9

☐B. 15

☐C. 21

☐D. 25

7- The sum of six different negative integers is \(-70\). If the smallest of these integers is\( -15\), what is the largest possible value of one of the other five integers?

☐A. \(-5\)

☐B. \(-2\)

☐C. \(-1\)

☐D. 0

8- What is the ratio of the minimum value to the maximum value of the following function?
\(-2 ≤ x ≤ 3? f(x) = -3x + 1\)

☐A. \(\frac{7}{8}\)

☐B. \(-\frac{8}{7}\)

☐C. \(-\frac{7}{8}\)

☐D. \(\frac{8}{7}\)

9-

Based on the graph, what’s the maximum ratio of woman to man in the four cities?

☐A. 0.98

☐B. 0.97

☐C. 0.96

☐D. 0.95

10-

Based on the graph, what’s the ratio of percentage of men in city \(A\) to percentage of women in city \(C\)?

☐A. 0.9

☐B. 0.95

☐C. 1

☐D. 1.05

Best PSAT Math Prep Resource for 2020

Answers:

1- A
Percent of cities in the type of pollution A:\( \frac{6}{10}×100=60\%\)
Percent of cities in the type of pollution C: \(\frac{4}{10}×100=40\%\)
Percent of cities in the type of pollution E:\( \frac{9}{10}×100=90\%\)

2- A
Let the number of cities should be added to type of pollutions \(B\) be \(x\). Then:
\(\frac{x+3}{8}=0.625→x+3=8×0.625→x+3=5→x=2\)

3- C
Let’s find the mean (average), mode and median of the number of cities for each type of pollution.
Number of cities for each type of pollution: \(6, 3, 4, 9, 8\)
\(Mean (Average) = \frac{sum \space of \space terms}{number \space of \space terms}=\frac{6+3+4+9+8}{5}=\frac{30}{5}=6\)
Median is the number in the middle. To find median, first list numbers in order from smallest to largest.
\(3, 4, 6, 8, 9\)
Median of the data is 6.
Mode is the number which appears most often in a set of numbers. Therefore, there is no mode in the set of numbers.
Median = Mean, then, a=c

4- C
The ratio of boy 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.
\(44 ÷ 11 = 4 ⇒ 4 × 4 = 16\)
There are 16 boys and \(28 (44 – 16)\) girls. So, 12 more boys should be enrolled to make the ratio \(1:1\)

5- A
\(AB = 12\) And\( AC = 5\)
\( BC = \sqrt {(12^2+5^2 )}=\sqrt{(144+25)}=\sqrt {169}=13 \)
Perimeter \(= 5 + 12 + 13 = 30\)
\(\frac{5×12}{2}=5×6=30\)
In this case, the ratio of the perimeter of the triangle to its area is: \(\frac{30}{30} = 1\)
If the sides \(AB\) and \(AC\) become twice longer, then:
\( AB=24 \space And \space AC=10\)
\(BC=\sqrt{(24^2+10^2 )}=\sqrt{(576+100)}=\sqrt {676}=26\)
Perimeter\( = 26 + 24 + 10 = 60\)
\(\frac{10×24}{2}=10×12=120\)
In this case the ratio of the perimeter of the triangle to its area is:
\(\frac{60}{120}=\frac{1}{2}\)

6- D
The capacity of a red box is \(20\%\) bigger than the capacity of a blue box and it can hold 30 books. Therefore, we want to find a number that \(20\%\) bigger than that number is 30. Let \(x\) be that number. Then:
\( 1.20 × x = 30\)
Divide both sides of the equation by 1.2. Then:
\(x = \frac{30}{1.20}=25\)

7- A
The smallest number is \(-15\). To find the largest possible value of one of the other five integers, we need to choose the smallest possible integers for four of them. Let \(x\) be the largest number. Then:
\(-70 = (-15) + (-14) + (-13) + (-12) + (-11) + x→-70 =- 65 + x
→x = -70 + 65 = -5\)

8- B
Since \(f(x)\) is linear function with a negative slop, then when \(x=-2, f(x)\) is maximum and when \(x=3, f(x)\) is minimum. Then the ratio of the minimum value to the maximum value of the function is:
\(\frac{f(3)}{f(-2)}=\frac{-3(3)+1}{-3(-2)+1}=\frac{- 8}{7}\)

9- B
Ratio of women to men in city A: \(\frac{570}{600}=0.95\)
Ratio of women to men in city B: \(\frac{291}{300}=0.97\)
Ratio of women to men in city C: \(\frac{667}{700}=0.95\)
Ratio of women to men in city D: \(\frac{528}{550} =0.96\)

10- D
Percentage of men in city \(A = \frac{600}{1170}×100 = 51.28 \% \)
Percentage of women in city \(C =
\frac{665}{1365} ×100 = 48.72\%\)
Percentage of men in city \(A\) to percentage of women in city \(C\)
\(\frac{51.28}{48.72} = 1.05\)

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