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

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

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New post 09 Aug 2019, 02:37
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

68% (00:50) correct 32% (01:02) wrong based on 36 sessions

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

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New post 09 Aug 2019, 02:53
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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New post 09 Aug 2019, 02:59
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

A is the answer
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Is x^2 > y^2 ? (1) x > y > 0 (2) x > 1 > y  [#permalink]

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New post 09 Aug 2019, 06:31
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

Answer is (A)

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Is x^2 > y^2 ? (1) x > y > 0 (2) x > 1 > y   [#permalink] 09 Aug 2019, 06:31
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