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24 Aug 2021, 08:00
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A
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Difficulty:
75%
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Question Stats:
52% (02:12) correct
48%
(02:09)
wrong
based on 218
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If x ≠ 0, is x < 1 ?
(1) \(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\).
(2) \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\).
M36-16
(1) \(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\).
(2) \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\).
M36-16
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28 Sep 2021, 04:20
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Bunuel
Official Solution:
If \(x ≠ 0\), is \(x < 1\) ?
(1) \(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\)
If \(x\) were a positive number, then \(\frac{1}{x}\) would be positive, but both \(-x^2\) and \(-x^3\) would be negative. This case is not possible because \(\frac{1}{x}\) would be the largest number in the list and thus cannot be the median. Therefore, \(x\) must be negative. Sufficient.
(2) \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\)
We can have the following cases:
A. \(-x \leq -\frac{1}{x^2} \leq x^5\). Multiply all three parts by \(x^2\) (we can safely do that since \(x^2\) will be positive): \(-x^3 \leq -1 \leq x^7\). From \(-x^3 \leq -1\) we get \(1 \leq x^3\) and finally \(1 \leq x\).
B. \(x^5 \leq -\frac{1}{x^2} \leq -x\). Multiply all three parts by \(x^2\) (we can safely do that since \(x^2\) will be positive): \(x^7 \leq -1 \leq -x^3\). From \(-1 \leq -x^3\) we get \(x^3 \leq 1\) and finally \(x \leq 1\).
Two different answers. Not sufficient.
Answer: A
General Discussion
24 Aug 2021, 08:59
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S1: assuming x to be +ve, no way a +ve number can be the median along with 2 other -ve numbers. Thus, x has to be < 0 . Suff.
S2:
take x as 2, we have -1/4 is the median of -1/4, -2, 32 - true.
take x as -2, we have -1/4 is the median of -1/4, 2, -32 - also true.
hence S2 is insuff.
Option A
S2:
take x as 2, we have -1/4 is the median of -1/4, -2, 32 - true.
take x as -2, we have -1/4 is the median of -1/4, 2, -32 - also true.
hence S2 is insuff.
Option A
