Top 10 SSAT Middle-Level Math Practice Questions

B. \(234\) cm\(^2\)

C. \(216\) cm\(^2\)

D. \(130\) cm\(^2\)

E. \(108\) cm\(^2\)

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Answers:

2- D
\(2×0.4=0.8\), all choices provided are equal to \(0.8\) except option D. \(\frac{5}{15}×3=1\)

3- C
Sara has M books. Mary has \(6\) more books than Sara. Then, Mary has \(M+6 \) books. If Mary gives Sara \(4\) books, Mary will have: \(M+6-4=M+2\)

4- E
If \(\frac{x}{2}=30\), then \(\frac{3x}{2}\) is \(3\) times \(\frac{x}{2}\). The answer is \(90\).

5- D
\(\frac{1}{6}\) of \(40\) is \(6.66\). Let’s review the choices provided:
A. \( 0.3×6=1.8\)
B. \( 0.3×5=1.5\)
C \( 0.2×30=6\)
D \( 0.2×35=7\)
E \( 0.2×39.5=7.9\)
Option D is the closest to \(6.66\)

6- B
The diagonal of the square is \(8\). Let \(x\) be the side.
Use Pythagorean Theorem: \(a^2+b^2=c^2\)
\(x^2 +x^2 = 8^2 ⇒ 2x^2 = 8^2 ⇒ 2x^2 = 64 ⇒x^2 = 32 ⇒x= \sqrt{32}\)
The area of the square is: \(\sqrt{32}×\sqrt{32}=32\)

7- B
The sum of supplement angles is \(180\). Let \(x \) be that angle. Therefore, \(x+5x=180\)
\(6x=180\), divide both sides by \(6: x=30\)

8- B
Use the formula for Percent of Change: \(\frac{New \ Value-Old \ Value}{Old Value}×100\% \)
\(\frac{28-44}{44}×100\%≈-36\%\) (negative sign here means that the new price is less than old price)

9- B
One liter \(=1,000 \) cm\(^3 → 6\) liters \(=6,000\) cm\(^3 \), \(6,000=25×5×h→h=\frac{6,000}{125}=48\) cm

10- A
The perimeter of the trapezoid is \(64\).
Therefore, the missing side (height) is \(= 64-18- 12-14 = 20\)
Area of the trapezoid: A \(= \frac{1}{2} h (b_{1}+b_{2 })= \frac{1}{2} (10) (12 + 14)=260\)

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$64.99 Original price was: $64.99.$36.99Current price is: $36.99.

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