10 Most Common SHSAT Math Questions

D
Simplify.
\(6x^2 y^3 (2x^2 y)^3= 6x^2 y^3 (8x^6 y^3 ) = 48x^8 y^6\)

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 \(= 2 (length + width) = 2 (8 + 6) = 2 (14) = 28\)

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 \(x\).
\((\frac{3}{2}) \frac{2}{3} x= \frac{1}{6} (\frac{3}{2}) {\Rightarrow} x= \frac{3}{12}=\frac{1}{4}\)

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

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
\(\frac{2}{3} < 67\% < 0.68 < \frac{4}{5}\ \)

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 \(= \frac{(4375)}{50}= 87.5\)

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

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. \(9 \div 1.5 = 6\)

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.

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

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