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m02 q 19

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Manager
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m02 q 19 [#permalink]

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New post 17 Dec 2011, 20:26
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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Re: m02 q 19 [#permalink]

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New post 26 Dec 2011, 21:22
1/2, -1/2 do no satisfy condition 1.
both condition need to be satisfied
Re: m02 q 19   [#permalink] 26 Dec 2011, 21:22
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