Top 10 ISEE Upper-Level Math Practice Questions

C.

D.

9- Solve.
\(|9 – (12 ÷ | 2 – 5 |)| =?\)

A. 9

B. \(-6\)

C. 5

D. \(-5\)

10- Solve: \( 2 – 10 ÷ (4^2 ÷ 2) =?\)

A. 6

B.\(\frac{3}{4}\)

C. \(-1\)

D. \(-2\)

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Answers:

2- B
\(\frac{240}{8} < x < \frac{320}{8}\)
\(30 < x < 40\)
Then:
Only choice b is correct

3- A
\((4.2 + 4.3 + 4.5) x = x\)
\(13x = x\)
Then \(x = 0\)

4- D
For sum of 6: (1 & 5) and (5 & 1), (2 & 4) and (4 & 2), (3 & 3), therefore we have 5 options.
For sum of 9: (3 & 6) and (6 & 3), (4 & 5) and (5 & 4), we have 4 options.
To get a sum of 6 or 9 for two dice: \(5 + 4 = 9\)
Since, we have \(6 × 6 = 36\) total number of options, the probability of getting a sum of 6 and 9 is 9 out of 36 or \(\frac{9}{36}=\frac{1}{4}\).

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5- C
Simplify:
\(\frac{\frac{1}{2}-\frac{x + 5}{4}}{\frac{x^2}{2}-\frac{5}{2}}=\frac{\frac{1}{2}-\frac{x + 5}{4}}{\frac{x^2 – 5}{2}}=\frac{2(\frac{1}{2}-\frac{x + 5}{4})}{x^2 – 5}\)
⇒Simplify:
\(\frac{1}{2} – \frac{x + 5}{4}=\frac{- x – 3}{4}\)
then:
\(\frac{2(\frac{-x – 3}{4})}{x^2 – 5}=\frac{\frac{-x – 3}{2}}{x^2 – 5}=\frac{- x – 3}{2(x^2 – 5)}=\frac{- x – 3}{2x^2 – 10}\)

6- B
\(x_{1,2} =\frac{-b ± \sqrt{b^2-4ac}}{2a}\)
\(ax^2 + bx + c = 0\)
\(4x^2 + 14x + 6 = 0\)
⇒ then: \(a = 4, b = 14\) and \(c = 6\)
\(x = \frac{-14 + \sqrt {14^2 – 4.4.6}}{2.4} = – \frac{1}{2}\)
\(x = \frac{-14 – \sqrt {14^2 – 4.4.6}}{2.4}= – 3\)

7- D
Use FOIL (First, Out, In, Last) method.
\((x + 7)(x + 5) = x^2 + 5x + 7x + 35 = x^2 + 12x + 3\)

8- D
Solve for \(x\).
\(-2≤2x-4<8\)⇒ (add 4 all sides) \(-2+4≤2x-4+4<8+4\)
⇒ \( 2≤2x<12\)
⇒ (divide all sides by 2) \(1 ≤ x < 6\)
\(x\) is between 1 and 6. Choice D represents this inequality.

9- A
\(|9 – (12 ÷ | 2 – 5 |)| = |(9-(12÷|-3|))|=|9-(12÷3)|=|9-4|=|5|=5\)

10- C
\(2 – 10 ÷ (4^2 ÷ 2) = 2 – 10 ÷ (16 ÷ 2) = 2 – 10 ÷ (8) = – 8 ÷ 8 = \ – 1\)

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