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29 Apr 2016, 03:57
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Question Stats:
52% (02:00) correct
48%
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wrong
based on 192
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If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y) > 0?
(1) x > y
(2) |x| > |y|
(1) x > y
(2) |x| > |y|
raarun
Joined: 01 Jul 2010
Last visit: 16 Mar 2018
Posts: 51
Given Kudos: 6
Location: India
Schools: Mannheim"18 Cranfield '18 WBS '19 Erasmus '18
GMAT 1: 660 Q43 V38
GPA: 3.4
29 Apr 2016, 08:30
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The question basically asks are both x+y and x-y positive or negative?
(1) x > y
(2) |x| > |y|
Stmnt 1:
x-y> 0 => x-y is positive
says nothing about x+y which can either be positive or negative
Insufficient
Stmnt 2
\(\sqrt{x^2}>\sqrt{y^2}\)
x^2-y^2 > 0
(x+y)(x-y) > 0
so both x+y and x-y can be positive or
both x+y and x-y can be negative
This means that x+y /x-y will always be positive and greater than 0
Sufficient
Ans B
(1) x > y
(2) |x| > |y|
Stmnt 1:
x-y> 0 => x-y is positive
says nothing about x+y which can either be positive or negative
Insufficient
Stmnt 2
\(\sqrt{x^2}>\sqrt{y^2}\)
x^2-y^2 > 0
(x+y)(x-y) > 0
so both x+y and x-y can be positive or
both x+y and x-y can be negative
This means that x+y /x-y will always be positive and greater than 0
Sufficient
Ans B
General Discussion
22 Aug 2018, 23:26
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(x + y)/(x - y) = (x + y)(x - y) / (x - y) (x - y) = (x^2 -y^2)/(x - y)^2
we know x is not equal to y , so (x - y)^2 is always > 0
Therefore x^2 - y^2 > 0
So , x^2 > y^2
Hence, |x| > |y|.
Ans- B
we know x is not equal to y , so (x - y)^2 is always > 0
Therefore x^2 - y^2 > 0
So , x^2 > y^2
Hence, |x| > |y|.
Ans- B
