m02 q 19 : Retired Discussions [Locked]
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# m02 q 19

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Manager
Joined: 13 May 2010
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17 Dec 2011, 19:26
Is $$x \gt y^2$$ ?

1. $$x \gt y + 5$$
2. $$x^2 - y^2 = 0$$

OA: c

Here is what the explanation says -

Statement (1) by itself is insufficient. Consider $$(x,y)=(5, -1)$$ or $$(5, -3)$$ .

Statement (2) by itself is insufficient. Consider $$(x,y)=(\frac{1}{2}, \frac{1}{2})$$ or $$(2, 2)$$ .

Statements (1) and (2) combined are sufficient. The absolute values of $$x$$ and $$y$$ are equal and $$x \gt y$$ , implying that $$x$$ is positive and $$y$$ is negative. Therefore, $$x \gt y^2$$ is false.

My point is why can't we consider x= 1/2 and y= -1/2 as the test case for combined case. If we consider that then even the combined statements case should be insufficient. Please suggest me if I am wrong.
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26 Dec 2011, 20:22
1/2, -1/2 do no satisfy condition 1.
both condition need to be satisfied
Re: m02 q 19   [#permalink] 26 Dec 2011, 20:22
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# m02 q 19

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