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Is x > y?

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Is x > y? [#permalink] New post 13 Feb 2013, 09:42
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Question Stats:

80% (02:01) correct 20% (00:23) wrong based on 1 sessions
Is x > y?

(1) x^2 > y

(2) x – |y| > 0

[Reveal] Spoiler:
My understtanding: All statement 1 tells us for sure is x is not 0
Statement 2 we can break into two staments

x>y
OR
x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 13 Feb 2013, 09:50, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Is x > y? [#permalink] New post 13 Feb 2013, 09:58
manimgoindowndown wrote:
Is x > y?

(1) x^2 > y

(2) x – |y| > 0

[Reveal] Spoiler:
My understtanding: All statement 1 tells us for sure is x is not 0
Statement 2 we can break into two staments

x>y
OR
x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?



Yes u got it right way! but u gotta complete it buddy :)

ya definitely S1 is not sufficent

But S2 is a trap.
x>y
OR
x>-y

assume numbers for x and y in this case
let x=4 and y=3 so 4>3
like wise x=4 and y=-3 this also 4>-3

So x>|y| always will be x>y

Hope i helped u :)
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Re: Is x > y? [#permalink] New post 15 Feb 2013, 02:35
From F.S 1, we can have x=-5,y=-3. Thus , x^2>y, but x<y. Hence the answer is NO. Now, for x=5,y=3, x^2>y and x>y. Here the answer is a YES. Thus not sufficient.

Fro F.S 2, x-mod(y)>0. Thus, x>mod(y). Whatever be the sign of y, x>y(always). Sufficient.

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Re: Is x > y? [#permalink] New post 15 Feb 2013, 03:16
manimgoindowndown wrote:
Is x > y?

(1) x^2 > y

(2) x – |y| > 0

[Reveal] Spoiler:
My understtanding: All statement 1 tells us for sure is x is not 0
Statement 2 we can break into two staments

x>y
OR
x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?


From 2, we get that x - |y| is positive.

Since |y| is postive, this means that x is also positive.

So, if "y" is negative, it automatically means that x is greater than y.

If "y" is positive,

x - y still gives a postive number. Hence x is again greater than y.

Answer is hence B.
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Re: Is x > y?   [#permalink] 15 Feb 2013, 03:16
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