10 Most Common FTCE Math Questions
Preparing for the FTCE Math test? Want a preview of the most common mathematics questions on the FTCE Math test? If so, then you are in the right place. For additional educational resources, visit the U.S. Department of Education website.
The mathematics section of FTCE can be a challenging area for many test-takers, but with enough patience, it can be easy and even enjoyable! For additional educational resources, visit the U.S. Department of Education website.
Preparing for the FTCE Math test can be a nerve-wracking experience. Learning more about what you’re going to see when you take the FTCE can help to reduce those pre-test jitters. Here’s your chance to review the 10 most common FTCE Math questions to help you know what to expect and what to practice most. Try these 10 most common FTCE Math questions to hone your mathematical skills and to see if your math skills are up to date on what’s being asked on the exam, or if you still need more practice. For additional educational resources, visit the U.S. Department of Education website.
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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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:
1- D
Simplify.
\(6x^2 y^3 (2x^2 y)^3= 6x^2 y^3 (8x^6 y^3 ) = 48x^8 y^6\)
2- 28 For education statistics and research, visit the National Center for Education Statistics.
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 For education statistics and research, visit the National Center for Education Statistics.
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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