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Difficulty:
55%
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Question Stats:
59% (02:01) correct
41%
(02:22)
wrong
based on 133
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If x is a non-zero number, is x^5 < x^7?
(1) x < x^2
(2) 1/x >1/x^2
(1) x < x^2
(2) 1/x >1/x^2
This is a question from e-gmat and i am not convinced with the way they have solved this question.
firas92
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23 Apr 2020, 13:07
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\(x^5 < x^7\) is true in the ranges \(-1<x<0\) and \(x>1\)
So we must find out if \(x\) lies in one of these two ranges
(1) \(x < x^2\)
\(x(x-1)>0\)
If \(x>0\), we have \(x-1>0\) or \(x>1\) (in this case \(x^5<x^7\))
If \(x<0\), we have \(x-1<0\) or \(x<1\) so \(x\) may or may not be in the range \(-1<x<0\)
1 is not sufficient
(2) \(\frac{1}{x} >\frac{1}{x^2}\)
Since \(x\) is non zero, \(x^2\) must be positive so we can multiply both sides by \(x^2\) and keep the same inequality
\(\frac{x^2}{x}>\frac{x^2}{x^2}\)
\(x>1\) so yes, \(x^5<x^7\)
2 is Sufficient
Answer is (B)
Posted from my mobile device
So we must find out if \(x\) lies in one of these two ranges
(1) \(x < x^2\)
\(x(x-1)>0\)
If \(x>0\), we have \(x-1>0\) or \(x>1\) (in this case \(x^5<x^7\))
If \(x<0\), we have \(x-1<0\) or \(x<1\) so \(x\) may or may not be in the range \(-1<x<0\)
1 is not sufficient
(2) \(\frac{1}{x} >\frac{1}{x^2}\)
Since \(x\) is non zero, \(x^2\) must be positive so we can multiply both sides by \(x^2\) and keep the same inequality
\(\frac{x^2}{x}>\frac{x^2}{x^2}\)
\(x>1\) so yes, \(x^5<x^7\)
2 is Sufficient
Answer is (B)
Posted from my mobile device
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23 Apr 2020, 14:45
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UShukla
1) x can lie anywhere in the number line except 0< x< 1. Depending on the sign of x, x^5 could be greater or lower than x^7. for x =-2, x^5 =- -32, x^7 =-128, so x^5 > x^7. But for positive sigh, the opposite happens. not sufficient
2) 1/x > 1/x^2, or, x < x^2. Same info as 1. Not sufficient
E is the answer.
