10 Most Common FTCE Math Questions
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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