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If x > 0, is x > y? (1) |x - y| > |y| (2) |x - y| = 3

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If x > 0, is x > y? (1) |x - y| > |y| (2) |x - y| = 3  [#permalink]

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New post 09 Dec 2019, 01:35
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  55% (hard)

Question Stats:

65% (02:00) correct 35% (02:04) wrong based on 33 sessions

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If x > 0, is x > y? (1) |x - y| > |y| (2) |x - y| = 3  [#permalink]

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New post 09 Dec 2019, 05:16
If x > 0, is x > y?

STATEMENT (1) |x - y| > |y|
squaring both sides we get,
x^2+y^2-2xy > y^2
x^2-2xy > 0
x(x-2y) > 0
(x-2y) > 0 (since x>0 )

x>2y and x positive value
any value of y will always be less than x


is x > y? --YES
SUFFICIENT

STATEMENT (2) |x - y| = 3

suppose x = 5, y = 2
then x>y
is x > y? --answer is YES

if x = 2 , y = 5 then |x - y| = 3
is x > y? --NO

INSUFFICIENT

A is the answer
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If x > 0, is x > y? (1) |x - y| > |y| (2) |x - y| = 3   [#permalink] 09 Dec 2019, 05:16
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