TASC Math FREE Sample Practice Questions

3- B
To find the discount, multiply the number by \((100\%-\)rate of discount).
Therefore, for the first discount we get: \((D) (100\%-15\%) = (D) (0.85) = 0.85 D\)
For increase of \(10\%: (0.85 D) (100\%+10\%)=(0.85 D) (1.10)=0.93D =93\%\) of D

4- B
Use the formula for Percent of Change \(\frac{New \ Value-Old \ Value}{Old \ Value}×100\%\)
\(\frac{28-45}{45}×100\%=-37.7\%\) (Negative sign here means that the new price is less than the old price).

5- D
Some of prime numbers are: \(2,3,5,7,11,13\). Find the product of two consecutive prime numbers: \(2×3=6\) (not in the options),\(3×5=15\) (not in the options),\(5×7=35\) (bingo!),
\(7×11=77\) (not in the options)

6- C
Let’s compare each fraction:\(\frac{1}{7} < \frac{3}{8} < \frac{5}{11} < \frac{5}{7}\). Only choice C provides the right order.

7- B
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: \(a^2+ b^2=c^2 \)
\(120^2+50^2=c^2 ⇒ 14400+2500= c^2 ⇒ 16900=c^2 ⇒ c=130 \)
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8- C
The ratio of boys 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 \(4. 55÷11=5 ⇒ 4×5=20\). There are \(20\) boys and \(35 (55-20) \)girls. So, \(15\) more boys should be enrolled to make the ratio \(1:1\)

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9- B
The question is this: \(530.20\) is what percent of \(631\)? Use percent formula:
part\(=\frac{percent}{100}×\)whole. \(530.20=\frac{percent}{100}×631 ⇒ 530.20= \frac{percent ×631}{100} ⇒ 53020 =\)percent \(×631 ⇒\) percent\(=\frac{53020}{631}=84.02≅84\) .
\(530.20\) is \(84\%\) of \(631\). Therefore, the discount is: \(100\%-84\%=16\%\)

10- B
If the score of Mia was \(40\), therefore the score of Ava is \(20\). Since the score of Emma was half as that of Ava, therefore, the score of Emma is \(10\).

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