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10 Aug 2020, 00:51
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (01:48) correct
38%
(02:02)
wrong
based on 152
sessions
History
Date
Time
Result
Not Attempted Yet
10 Aug 2020, 04:02
Kudos
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Given:
Nothing
To Find:
Is \(\frac{1}{(y-x)} > x-y\)
Pre Process:
\(\frac{1}{(y-x)} > x-y\)
\(\frac{1}{(y-x)} - (x-y) >0\)
\(\frac{1}{(y-x)} + (y-x) >0\)
\(\frac{1 + (y-x)^2}{(y-x)(} >0\)
Since numerator is always positive, we need to find whether denominator is positive.
\(y-x > 0\)
y >x ?
Now coming to the statements:
1) \(y > x.\)
Just what we need. So given expression is positive.
1) is Sufficient.
2) \(|x-y| > 0.\)
The only thing this tells us is that x is not equal to y.
But no information about \(y>x.\)
2) is not Sufficient.
Hence correct option is A).
Nothing
To Find:
Is \(\frac{1}{(y-x)} > x-y\)
Pre Process:
\(\frac{1}{(y-x)} > x-y\)
\(\frac{1}{(y-x)} - (x-y) >0\)
\(\frac{1}{(y-x)} + (y-x) >0\)
\(\frac{1 + (y-x)^2}{(y-x)(} >0\)
Since numerator is always positive, we need to find whether denominator is positive.
\(y-x > 0\)
y >x ?
Now coming to the statements:
1) \(y > x.\)
Just what we need. So given expression is positive.
1) is Sufficient.
2) \(|x-y| > 0.\)
The only thing this tells us is that x is not equal to y.
But no information about \(y>x.\)
2) is not Sufficient.
Hence correct option is A).
ArunSharma12
Joined: 25 Oct 2015
Last visit: 20 Jul 2022
Posts: 513
Own Kudos:
Given Kudos: 74
Location: India
Schools: IIML-IPMX '22 (A)
GMAT 1: 650 Q48 V31
GMAT 2: 720 Q49 V38 (Online)
GPA: 4
Products:
10 Aug 2020, 04:05
Kudos
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Quote:
Given:
\(\frac{1}{(y − x)} > x − y \)
\(\frac{1}{(y − x)} + y - x > 0 \)
\(\frac{1 + (y − x)(y - x)}{y - x} > 0 \)
\(\frac{1 + (y − x)^2 }{ y -x} > 0 \)
statement 1: y > x
(y - x)^2 is a non-negative value.
numerator will always be a positive value.
as y > x, denominator will be positive irrespective of the sign of y and x.
sufficient
statement 2: |x - y| > 0
\(x \ne y\), but we don't know whether x is greater than y or vice-versa?
not sufficient
Ans: A
