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# Is x > y^2? 1. x > y + 5 2. x^2 - y^2 = 0

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Manager
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Is x > y^2? 1. x > y + 5 2. x^2 - y^2 = 0 [#permalink]

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02 Sep 2008, 07:41
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Question Stats:

35% (02:07) correct 65% (01:34) wrong based on 174 sessions

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Is x > y^2?

(1) x > y + 5
(2) x^2 - y^2 = 0

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-y-2-1-x-y-5-2-x-2-y-97142.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Jun 2012, 07:35, edited 1 time in total.
Edited the question and added the OA. Topic is locked.

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Manager
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Re: DS : m02 #19 [#permalink]

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02 Sep 2008, 08:08
St 2 : x^2 = y^2 Hence x cannot be > y^2.
St.1: x> y+5. Multiple values can be substituted for x, y.-- Insufficient
Therefore, ans should be B

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Re: DS : m02 #19 [#permalink]

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02 Sep 2008, 08:23
KASSALMD wrote:
St 2 : x^2 = y^2 Hence x cannot be > y^2.
St.1: x> y+5. Multiple values can be substituted for x, y.-- Insufficient
Therefore, ans should be B

x=1/2 y=1/2 x>y^2 true

x=2 y=2 x>y^2 false
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Re: DS : m02 #19 [#permalink]

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02 Sep 2008, 11:49
Guys,

Its not B.

For Statement 2:
x=1/2 y=1/2 x>y^2 true

x=2 y=2 x>y^2 false

combine

x and y must have opposite sign because x>y+5

x>-x+5
x>2.5

Sufficient

Should be C.
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Re: DS : m02 #19 [#permalink]

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02 Sep 2008, 12:04
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Real good one! Thanks for the post.

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Re: DS : m02 #19 [#permalink]

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02 Sep 2008, 14:21
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x2suresh wrote:
Guys,

Its not B.

For Statement 2:
x=1/2 y=1/2 x>y^2 true

x=2 y=2 x>y^2 false

combine

x and y must have opposite sign because x>y+5

x>-x+5
x>2.5

Sufficient

Should be C.

thats the trick.
good stuff

OA is C

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Re: Is x > y^2? 1. x > y + 5 2. x^2 - y^2 = 0 struggling [#permalink]

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22 Jun 2012, 05:45
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Thnx For the explanation

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Manager
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Re: Is x > y^2? 1. x > y + 5 2. x^2 - y^2 = 0 struggling [#permalink]

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22 Jun 2012, 07:30
seems like answer is C, as it is not mentioned they are integers!
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Re: Is x > y^2? 1. x > y + 5 2. x^2 - y^2 = 0 [#permalink]

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22 Jun 2012, 07:35
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Expert's post
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Is $$x>y^2$$?

(1) $$x>y+5$$ --> $$x-y>5$$. Clearly insufficient, for example: if $$x=1$$ and $$y=-10$$ then the answer is NO, but if $$x=10$$ and $$y=1$$ then the answer is YES. Two different answers, hence not sufficient.

(2) $$x^2-y^2=0$$ --> $$(x-y)(x+y)=0$$ --> so either $$x-y=0$$ or $$x+y=0$$. Also insufficient: if $$x=1$$ and $$y=1$$, then answer is NO, buy if $$x=\frac{1}{2}$$ and $$y=\frac{1}{2}$$, then the answer is YES. Two different answers, hence not sufficient.

(1)+(2) As from (1) $$x-y>5\neq{0}$$, then from (2) must be true that $$x+y=0$$ --> so $$x=-y$$ --> substitute $$x$$ in (1) --> $$-y-y>5$$ --> $$y<-\frac{5}{2}<0$$, as $$x=-y$$, then $$x>\frac{5}{2}>0$$, so $$y^2$$ (or which is the same $$x^2$$) will always be more than $$x$$, thus the answer to the question "Is $$x>y^2$$" is NO. Sufficient.

To elaborate more as $$x=-y>0$$, the only chance for $$x>y^2$$ to hold true (or which is the same for $$x>x^2$$ to hold true) would be if $$x$$ is fraction ($$0<x<1$$). For example if $$x=\frac{1}{2}$$ and $$y=-\frac{1}{2}$$ then $$x=\frac{1}{2}>y^2=\frac{1}{4}$$. But the fact that $$x>\frac{5}{2}>0$$ rules out this option.

Hope it's clear.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-y-2-1-x-y-5-2-x-2-y-97142.html
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Re: Is x > y^2? 1. x > y + 5 2. x^2 - y^2 = 0   [#permalink] 22 Jun 2012, 07:35
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