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Is x > y^2 ?

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Is x > y^2 ?

by vinni.k » Fri Dec 28, 2012 11:37 pm
Is x > y^2 ?
(1) x > y + 5
(2) x^2 - y^2 = 0

Answer is C

Please explain with the help of examples

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by Anurag@Gurome » Sat Dec 29, 2012 10:49 am
vinni.k wrote:Is x > y^2 ?
(1) x > y + 5
(2) x^2 - y^2 = 0
Statement 1: Consider the following two examples,
  • x = 0 and y = -10 ---> x < y²
    x = 6 and y = 0 ---> x > y²
Not sufficient

Statement 2: Consider the following two examples,
  • x = 2 and y = 2 ---> x < y²
    x = 1/2 and y = 1/2 ---> x > y²
Not sufficient

1 & 2 Together: From statement 2, we can conclude that either x = y or x = -y
Now if x = y,
  • x > y + 5 --> x > x + 5 --> 0 > 5 --> Not possible
But if x = -y,
  • x > y + 5 --> x > -x + 5 --> 2x > 5 --> x > 2.5 > 0
As x > 2.5 and x² = y², x will be always less than y².

Sufficient

The correct answer is C.
Last edited by Anurag@Gurome on Mon Dec 31, 2012 2:24 am, edited 1 time in total.
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by sana.noor » Sat Dec 29, 2012 11:21 pm
Isn't statement 2 enough to answer...as x^2-y^2=0...so either x=y or x=-y???
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by vinni.k » Sun Dec 30, 2012 9:32 pm
Anurag@Gurome wrote:
Statement 2: Consider the following two examples,
  • x = 2 and y = 2 ---> x < y²
    x = 0 and y = 0 ---> x = y²
Not sufficient
Anurag, thanks for your reply. But i still believe that statement (2) is sufficient to answer the question. The question is asking "Is x > y² ?
After going through your examples that say x < y² and x = y². In both cases, the answer is "no" and we can select B.
Can you please put some more light on it. I am getting more confused now.

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by vinni.k » Sun Dec 30, 2012 9:47 pm
vinni.k wrote:
Anurag@Gurome wrote:
Statement 2: Consider the following two examples,
  • x = 2 and y = 2 ---> x < y²
    x = 0 and y = 0 ---> x = y²
Not sufficient
I have just realized that if we take x = 1/2 and y = 1/2 instead of x = 0 and y = 0, then the answer is yes. So, one is yes and another is no for statement (2).

Thanks Anurag.

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by Anurag@Gurome » Mon Dec 31, 2012 2:28 am
Thanks for pointing it out vinni.
I've edited my reply.
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