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DS - X > Y^2?

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Senior Manager
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DS - X > Y^2? [#permalink] New post 04 Aug 2007, 11:13
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Is X > Y^2?

1) X > Y+5
2) X^2 - Y^2 = 0

Please explain. Thanks!
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 [#permalink] New post 04 Aug 2007, 11:31
looks E to me..cant tell if X or Y are positive or negative....
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Re: DS - X > Y^2? [#permalink] New post 04 Aug 2007, 14:29
leeye84 wrote:
Is X > Y^2?

1) X > Y+5
2) X^2 - Y^2 = 0

Please explain. Thanks!


Got C, hope it's right. I spend way more than 2 mins. what' OA?

1) x > y+5
Try x=7, y=1: 7>1 yes!
Try x=1, y=-5: 1>25 No!
INSUFFICIENT.

2) x^2 - y^2 = 0
This is same as |x| = |y|
This means that the sign doesn't matter.
In other word, (+/-)x = (+/-)y
Try x=1, x=-1 => (+/-)1 will not be greater than 1
Try x=0, 0 will not be greater than 0.
Try large number, say x=10
(+/-)10 will not ever be greater than 100
*However, if you plug in 0<x<1, then x>y^2
This means that only 0<x<1 works but not others.
INSUFFICIENT.

Together, we know that x-y > 5. This means that the difference between them is more than 5. From |x|=|y|, it tell us that x and y will always have different sign and their value is at least 2.5
This is actually sufficient to answer the question since x won't be between 0 and 1.

*I edited this post.*
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Re: DS - X > Y^2? [#permalink] New post 04 Aug 2007, 19:57
leeye84 wrote:
Is X > Y^2?

1) X > Y+5
2) X^2 - Y^2 = 0

Please explain. Thanks!


C. x is positive and y is negative.

x = lyl

suppose y = - 3, x = 3, which is > y + 5.
suppose y = - 4, x = 4, which is > y + 5.

y must be less than 2.5. or y cannot be 2.5 or grater than -2.5.
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 [#permalink] New post 05 Aug 2007, 03:08
I go with C and answer is No.
1. x>y+5. Insufficient

2. lxl=lyl.
Insufficient as both can be 0 or any other integer.

C. x has to be postive and Y has to be -ve. and the x> 2.5

thus x will always be less than y*2.
  [#permalink] 05 Aug 2007, 03:08
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