10 Most Common CBEST Math Questions

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10 Sample CBEST Math Practice Questions

1- Simplify \(6x^2 y^3 (2x^2 y)^3= \)

A. \(12x^4 y^6\)

B. \(12x^8 y^6\)

C. \(48x^4 y^6\)

D. \(48x^8 y^6\)

2- The diagonal of a rectangle is \(10\) inches long and the height of the rectangle is \(8\) inches. What is the perimeter of the rectangle in inches? ________

3- What is the value of \(x\) in the following equation?
\(\frac{2}{3} x+\frac{1}{6}= \frac{1}{3}\)

A. \(6\)

B. \(\frac{1}{2} \)

C. \(\frac{1}{3} \)

D. \(\frac{1}{4} \)

4- A card is drawn at random from a standard \(52\)–card deck, what is the probability that the card is of Hearts? (The deck includes \(13\) of each suit clubs, diamonds, hearts, and spades)

A. \(\frac{1}{3} \)

B. \(\frac{1}{6} \)

C. \(\frac{1}{52} \)

D. \(\frac{1}{4} \)

5- Which of the following shows the numbers in ascending order?
\(\frac{2}{3} , 0.68 , 67\% , \frac{4}{5}\)

A. \(67\%, 0.68, \frac{2}{3} , \frac{4}{5} \)

B. \(67\%, 0.68, \frac{4}{5} , \frac{2}{3} \)

C. \(0.68, 67\%, \frac{2}{3} , \frac{4}{5} \)

D. \(\frac{2}{3} , 67\%, 0.68, \frac{4}{5} \)

6- The mean of \(50\) test scores was calculated as \(88\). But, it turned out that one of the scores was misread as \(94\) but it was \(69\). What is the mean?

A. \(85\)

B. \(87\)

C. \(87.5\)

D. \(88.5\)

7- Two dice are thrown simultaneously, what is the probability of getting a sum of \(6\) or \(9\)?

A. \(\frac{1}{3} \)

B. \(\frac{1}{4} \)

C. \(\frac{1}{6} \)

D. \(\frac{1}{2} \)

8- Jason is \(9\) miles ahead of Joe running at \(5.5\) miles per hour and Joe is running at the speed of \(7\) miles per hour. How long does it take Joe to catch Jason?

A. \(3\) hours

B. \(4\) hours

C. \(6\) hours

D. \(8\) hours

9- \(55\) students took an exam and \(11\) of them failed. What percent of the students passed the exam?

A. \(40\%\)

B. \(60\%\)

C. \(80\%\)

D. \(20\%\)

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

A. \(15 cm^3\)

B. \(60 cm^3\)

C. \(90 cm^3\)

D. \(120 cm^3\)

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Answers:

2- 28

Let x be the width of the rectangle. Use Pythagorean Theorem:
\(a^2 + b^2 = c^2\)
\(x^2 + 8^2 = 10^2 {\Rightarrow} x^2 + 64 = 100 {\Rightarrow} x^2 = 100 – 64 = 36 ⇒ x = 6\)
Perimeter of the rectangle \(= 2 (length + width) = 2 (8 + 6) = 2 (14) = 28\)

3- D
Isolate and solve for \(x\).
\(\frac{2}{3} x+\frac{1}{6} = \frac{1}{3} {\Rightarrow} \frac{2}{3} x= \frac{1}{3} -\frac{1}{6} = \frac{1}{6} {\Rightarrow} \frac{2}{3} x= \frac{1}{6} \)
Multiply both sides by the reciprocal of the coefficient of \(x\).
\((\frac{3}{2}) \frac{2}{3} x= \frac{1}{6} (\frac{3}{2}) {\Rightarrow} x= \frac{3}{12}=\frac{1}{4}\)

4- D
The probability of choosing a Hearts is \(\frac{13}{52}=\frac{1}{4} \)

5- D
Change the numbers to decimal and then compare.
\(\frac{2}{3} = 0.666… \)
\(0.68 \)
\(67\% = 0.67\)
\(\frac{4}{5} = 0.80\)
Therefore
\(\frac{2}{3} < 67\% < 0.68 < \frac{4}{5}\)

6- C
average (mean)\( =\frac{(sum \space of \space terms)}{(number \space of \space terms)} {\Rightarrow} 88 = \frac{(sum \space of \space terms)}{50} {\Rightarrow} sum = 88 {\times} 50 = 4400\)
The difference of \(94\) and \(69\) is \(25\). Therefore, \(25\) should be subtracted from the sum.
\(4400 – 25 = 4375\)
\(mean = \frac{(sum of terms)}{(number of terms)} ⇒ mean = \frac{(4375)}{50}= 87.5\)

7- B
To get a sum of \(6\) for two dice, we can get \(5\) different options:
\((5, 1), (4, 2), (3, 3), (2, 4), (1, 5)\)
To get a sum of \(9\) for two dice, we can get \(4\) different options:
\((6, 3), (5, 4), (4, 5), (3, 6)\)
Therefore, there are \(9\) options to get the sum of \(6\) or \(9\).
Since we have \(6 × 6 = 36\) total options, the probability of getting a sum of \(6\) and \(9\) is \(9\) out of \(36\) or \(\frac{1}{4}\).

8- C
The distance between Jason and Joe is \(9\) miles. Jason running at \(5.5\) miles per hour and Joe is running at the speed of \(7\) miles per hour. Therefore, every hour the distance is \(1.5\) miles less. \(9 \div 1.5 = 6\)

9- C
The failing rate is \(11\) out of \(55 = \frac{11}{55} \)
Change the fraction to percent:
\( \frac{11}{55} {\times} 100\%=20\% \)
\(20\) percent of students failed. Therefore, \(80\) percent of students passed the exam.

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

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