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GMAT 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 Easy

If \( a, b, c, \) and \( d \) are integers and \( ab^2c^3d^4 > 0 \), which of the following must be positive?
I. \( a^2cd \)
II. \( bc^4d \)
III. \( a^3c^3d^2 \)

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Correct answer: C

\( b^2, d^4 \) are positive. So \( ac^3 > 0 \). Since \( c^3 \) has same sign as \( c \), \( ac > 0 \). \( a, c \) same sign.
I. \( a^2cd \). \( a^2 > 0 \). \( cd \)? \( c, d \) signs unknown. Uncertain.
II. \( bc^4d \). \( c^4 > 0 \). \( bd \)? Unknown. Uncertain.
III. \( a^3c^3d^2 = (ac)^3 d^2 \). \( ac > 0 \implies (ac)^3 > 0 \). \( d^2 > 0 \). Product positive. True.
The correct answer is C (III only).
Question 2 of 3 free Easy

\( w, x, y, \) and \( z \) are positive integers. If \( \frac{w}{x} < \frac{y}{z} < 1 \), what is the proper order of magnitude, increasing from left to right, of the following quantities:
\( \frac{x}{w}, \frac{z}{y}, \frac{x^2}{w^2}, \frac{xz}{wy}, \frac{x + z}{w + y}, 1 \) ?

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Correct answer: E

Let \(p=x/w\) and \(q=z/y\). Inverting the given inequality gives \(p>q>1\).

Also, \((x+z)/(w+y)=(wp+yq)/(w+y)\), a weighted average of \(p\) and \(q\), so it lies strictly between them. Since \(q>1\), multiplying gives \(pq>p\); and since \(p>q\), \(p^2>pq\).

Therefore the increasing order is
\[1<q<\frac{x+z}{w+y}<p<pq<p^2,\]
or \(1<z/y<(x+z)/(w+y)<x/w<xz/(wy)<x^2/w^2\). The correct answer is E.
Question 3 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 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 Hard

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

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