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Re: GMAT CLUB OLYMPICS: If x 0, is x < 1 ? [#permalink]
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I tried to explain everything in detail in the attached file.
My answe is A
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Re: GMAT CLUB OLYMPICS: If x 0, is x < 1 ? [#permalink]
Expert Reply
Bunuel wrote:
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


x ≠ 0

(1) \(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\).
\(-x^2\) is always negative, while \(\frac{1}{x}\) and \(-x^3\) will have opposite sign.
When x>0, \(\frac{1}{x}>0\), while \(-x^2\) and \(-x^3\) will be negative. This case is not possible.
Hence x<0.
Sufficient

(2) \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\).
When x>0, \(-\frac{1}{x^2}<0\), while \(x^5>0\) and \(-x<0\). Always true for x>1.
When x<0, \(-\frac{1}{x^2}<0\), while \(x^5<0\) and \(-x>0\). Always true for x<-1.
Hence, Insufficient


A
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Re: GMAT CLUB OLYMPICS: If x 0, is x < 1 ? [#permalink]
Expert Reply
Bunuel wrote:
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


x ≠ 0

(1) \(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\).
When x>0, \(\frac{1}{x}>0\), while \(-x^2\) and \(-x^3\) will be negative. This case is not possible.
Hence x<0.
Sufficient

(2) \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\).
\(-\frac{1}{x^2}\) is always negative, while \(-x\) and \(-x^5\) will have opposite sign.
Thus, the signs of three in increasing order would be (-, -, +), which is possible both when x>0 and when x<0.
Insufficient


A
GMAT Club Bot
Re: GMAT CLUB OLYMPICS: If x 0, is x < 1 ? [#permalink]
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