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Inequalities and Absolute Value Practice Questions

Practice Inequalities and Absolute Value questions with worked explanations and timing guidance for Quantitative Reasoning.

Five-question preview. Answer 3 now without an account.
Question 1 of 3 free Medium

If \( 3|3 - x| = 7 \), what is the product of all the possible values of \( x \)?

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Correct answer: E

When solving an absolute value equation, it helps to first isolate the absolute value expression:

\( 3|3 - x| = 7 \)
\( |3 - x| = \frac{7}{3} \)

When removing the absolute value bars, we need to keep in mind that the expression inside the absolute value bars \( (3 - x) \) could be positive or negative. Let's consider both possibilities:

When \( (3 - x) \) is positive:
\( (3 - x) = \frac{7}{3} \)
\( 3 - \frac{7}{3} = x \)
\( \frac{9}{3} - \frac{7}{3} = x \)
\( x = \frac{2}{3} \)

When \( (3 - x) \) is negative:
\( -(3 - x) = \frac{7}{3} \)
\( x - 3 = \frac{7}{3} \)
\( x = \frac{7}{3} + 3 \)
\( x = \frac{7}{3} + \frac{9}{3} \)
\( x = \frac{16}{3} \)
So, the two possible values for \( x \) are \( \frac{2}{3} \) and \( \frac{16}{3} \). The product of these values is \( \frac{32}{9} \).
The correct answer is E.
Question 2 of 3 free Medium

If \( |a| = \frac{1}{3} \) and \( |b| = \frac{2}{3} \), which of the following CANNOT be the result of \( a + b \)?

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Correct answer: D

The possible values for \( a \) and \( b \) are \( \pm \frac{1}{3} \) and \( \pm \frac{2}{3} \) respectively.
We can list the possible sums:

$$\begin{array}{|c|c|c|} \hline a & b & a+b \\ \hline \frac{1}{3} & \frac{2}{3} & 1 \\ \hline \frac{1}{3} & -\frac{2}{3} & -\frac{1}{3} \\ \hline -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \hline -\frac{1}{3} & -\frac{2}{3} & -1 \\ \hline \end{array}$$

Comparing with the options, \( \frac{2}{3} \) is NOT a possible sum.
The correct answer is D.
Question 3 of 3 free Hard

If \( |x - \frac{9}{2}| = \frac{5}{2} \), and if \( y \) is the median of a set of \( p \) consecutive integers, where \( p \) is odd, which of the following must be true?
I. \( xyp \) is odd
II. \( xy(p^2 + p) \) is even
III. \( x^2y^2p^2 \) is even

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Correct answer: A

Solve for \( x \): \( x - 4.5 = 2.5 \implies x = 7 \) OR \( x - 4.5 = -2.5 \implies x = 2 \). Therefore, \( x \) can be either odd (7) or even (2).

\( p \) is odd. \( y \) is the median of \( p \) consecutive integers, so \( y \) can be odd or even.

I. \( xyp \) is odd: This is not necessarily true. If \( x = 2 \), then \( xyp \) is even.

II. \( xy(p^2 + p) = xyp(p + 1) \): Since \( p \) is odd, \( p + 1 \) is even. The product contains an even factor, so it is always even. This must be true.

III. \( x^2y^2p^2 \) is even: This is not necessarily true. If \( x = 7 \), \( y \) is odd, and \( p \) is odd, then \( x^2y^2p^2 \) is odd.

Only statement II must be true.
Question 4 of 5 preview Medium

If \( \sqrt{(x + 4)^2} = 3 \), which of the following could be the value of \( x - 4 \)?

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Question 5 of 5 preview Medium

If \( x > y \), \( x^2 - 2xy + y^2 - 9 = 0 \), and \( x + y = 15 \), what is \( x \)?

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