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14 May 2018, 01:08
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GMAT CLUB'S FRESH QUESTION
If x ≠ 1, what is the value of \(\frac{-|x| - 1}{x - 1}\)?
(1) \(\sqrt{x^6} > x^3\)
(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\)
M36-06
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14 May 2018, 01:36
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Bunuel
GMAT CLUB'S FRESH QUESTION
If x ≠ 1, what is the value of \(\frac{-|x| - 1}{x - 1}\)?
(1) \(\sqrt{x^6} > x^3\)
(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\)
The data is sufficient if we get a unique value for the expression \(\frac{-|x| - 1}{x - 1}\)
A quick info about this expression. If x is negative then the expression will have a unique value, which is 1.
If x is 0, it will have a unique value, which is 1.
If x is positive it will have multiple values. For instance, if x = 2, the value of the expression is -3. If x = 3, the value of the expression -2.
So, at some level if we can establish that x is not positive, we will have a unique value. Else we will not have a unique value.
Statement 1: \(\sqrt{x^6} > x^3\)
\(\sqrt{x^6}\) is non negative for real x and will be |x^3|.
For instance \(\sqrt{2^6}\) = 8 and \(\sqrt{{(-2)^6}\) will also be 8.
But 2^3 will be 8 and (-2)^3 will be -8.
So, if we know that \(\sqrt{x^6} > x^3\), we can infer that x is negative.
If x is negative, the value of \(\frac{-|x| - 1}{x - 1}\) will be 1. Take any negative x and check it. Negative integer, negative non integer. It will work for all values.
So, statement 1 ALONE is sufficient.
Statement 2:\(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\)
We can rewrite the expression as |x| + 2x + 4|x| + 8x = -16
5|x| + 10x = -16.
The sum of the expression is -16, which is negative.
5|x| cannot be negative. So, 10x has to be negative if the sum is -16 => x has to be negative.
If x is negative, the expression \(\frac{-|x| - 1}{x - 1}\) has a unique value which is 1.
Statement 2 ALONE is sufficient.
Each statement is INDEPENDENTLY sufficient.
Choice D.
21 Sep 2021, 07:54
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Official Solution:
If \(x ≠ 1\), what is the value of \(\frac{-|x| - 1}{x - 1}\)?
Work on the stem before moving to the statements.
If \(x \leq 0\), then \(|x|=-x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). So, if \(x\) is 0 or negative then the value of the fraction is 1.
If \(x > 0\), then \(|x|=x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-x - 1}{x - 1}\). So, in this case the value of the fraction would depend on the value of \(x\).
(1) \(\sqrt{x^6} > x^3\). Take the square root from the left hand side: \(|x^3| > x^3\). For this to hold true, \(x\) must be negative, because if \(x\) is 0 or positive, then \(|x^3|\) would be equal to \(x^3\). From above, we know that if \(x < 0\), then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.
(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\). Notice that if \(x\) is positive, then \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = positive +positive+positive+positive=positive\), not negative number \(-1\). So, \(x\) cannot be positive. From above, we know that if \(x\) is not positive, then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.
Answer: D
If \(x ≠ 1\), what is the value of \(\frac{-|x| - 1}{x - 1}\)?
Work on the stem before moving to the statements.
If \(x \leq 0\), then \(|x|=-x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). So, if \(x\) is 0 or negative then the value of the fraction is 1.
If \(x > 0\), then \(|x|=x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-x - 1}{x - 1}\). So, in this case the value of the fraction would depend on the value of \(x\).
(1) \(\sqrt{x^6} > x^3\). Take the square root from the left hand side: \(|x^3| > x^3\). For this to hold true, \(x\) must be negative, because if \(x\) is 0 or positive, then \(|x^3|\) would be equal to \(x^3\). From above, we know that if \(x < 0\), then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.
(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\). Notice that if \(x\) is positive, then \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = positive +positive+positive+positive=positive\), not negative number \(-1\). So, \(x\) cannot be positive. From above, we know that if \(x\) is not positive, then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.
Answer: D
