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# Is x^2 > y^2 ? (1) x > y > 0 (2) x > 1 > y

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Math Expert
Joined: 02 Sep 2009
Posts: 64163
Is x^2 > y^2 ? (1) x > y > 0 (2) x > 1 > y  [#permalink]

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09 Aug 2019, 01:37
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Difficulty:

35% (medium)

Question Stats:

65% (00:59) correct 35% (01:13) wrong based on 49 sessions

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Is x^2 > y^2 ?

(1) x > y > 0
(2) x > 1 > y

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Joined: 26 Mar 2019
Posts: 154
Re: Is x^2 > y^2 ? (1) x > y > 0 (2) x > 1 > y  [#permalink]

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09 Aug 2019, 01:53
Value of any negative square is always bigger hence from statement 1 we know that x and y are negative so A is sufficient.

Statement two x is greater than 1 and y is lower than 1 so if we take 2 as x and -3 as y the statement is satisfied but the sqaure of both number would flip inequities. Hence statement 2 is not sufficient

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Re: Is x^2 > y^2 ? (1) x > y > 0 (2) x > 1 > y  [#permalink]

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09 Aug 2019, 01:59
Is $$x^2$$>$$y^2$$?

STATEMENT 1--x > y > 0
From this we know x and y----positive and x>y
so ,$$x^2$$>$$y^2$$
when x = 2 y = 1
$$x^2$$>$$y^2$$
when x = $$\frac{1}{2}$$ y = $$\frac{1}{3}$$
$$x^2$$>$$y^2$$
so , SUFFICIENT

STATEMENT 2--x > 1 > y
X>1 and y<1
when x =2 y = 0
then $$x^2$$>$$y^2$$ --YES
when x = 2 y = -4
then $$x^2$$<$$y^2$$ --NO
so INSUFFICIENT

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GMAT 1: 720 Q49 V40
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Is x^2 > y^2 ? (1) x > y > 0 (2) x > 1 > y  [#permalink]

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09 Aug 2019, 05:31
Is x^2>y^2 ?
--> Is (x-y)(x+y)>0 ?
--> Are both (x-y) and (x+y) of the same sign ?

(1) x > y > 0
We know from statement (1) as follows:
a. x > y --> x - y > 0 --> (x-y) is positive
b. x > 0 and y > 0 both positive --> (x+y) is positive
--> Both (x-y) and (x+y) are of the same sign. Thus, x^2>y^2 must be true
SUFFICIENT

(2) x > 1 > y
We know from statement (2) as follows:
a. x > y --> x - y > 0 --> (x-y) is positive
b. x>1 is positive. However, y<1 could be positive or negative of zero --> (x+y) could be positive or negative or zero
--> Both (x-y) and (x+y) could be of either the same sign or the opposite sign. Thus, x^2>y^2 could be either true of false.
NOT SUFFICIENT

+1 kudo if this solution is deemed good
Is x^2 > y^2 ? (1) x > y > 0 (2) x > 1 > y   [#permalink] 09 Aug 2019, 05:31