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Is x > y?

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Is x > y? [#permalink]

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New post 05 Jul 2011, 02:46
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Is x > y?

(1) x^2 > y

(2) x – |y| > 0
[Reveal] Spoiler: OA

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Re: Is x > y? [#permalink]

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New post 05 Jul 2011, 03:18
mojorising800 wrote:
can somebody please explain this.Looks a very easy question but i am confused about the OE given.
Is x > y?

(1) x^2 > y

(2) x – |y| > 0



Statement 1 is clearly not sufficient alone. Only if we know that x > x^2, will we get a definite answer. But x > x^2 only when x lies between 0 and 1 which is not given. So statement 1 is not sufficient.

Statement 2 tells you that x > |y|. This means x is greater than the absolute value of y. Then x has to be greater than y too.
Think about it: |y| is either equal to y (when y is +ve or 0) or it is greater than y (when y is negative). So if x is greater than |y|, it has to be greater than y too.

You can also take an example to understand this.
Say y = -4
Given x > |y| i.e. x > 4
Then x has to be greater than -4 too.
Say y = 4
Given x > |y| i.e. x > 4
In either case, x has to be greater than y.
So statement 2 alone is sufficient.
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Re: Is x > y? [#permalink]

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New post 05 Jul 2011, 03:49
mojorising800 wrote:
can somebody please explain this.Looks a very easy question but i am confused about the OE given.
Is x > y?

(1) x^2 > y

(2) x – |y| > 0



The best way is to test cases.

1) take x=-2, y=1 and x=2 , y=1. NS

2) rewrite it , so x>|y|
in this case x is positive , so x>0 , thus in any case of y whether positive or negative x>y.
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Re: Is x > y? [#permalink]

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New post 27 Aug 2016, 08:26
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Re: Is x > y?   [#permalink] 27 Aug 2016, 08:26
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