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31 Mar 2010, 15:34
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Difficulty:
35%
(medium)
Question Stats:
66% (01:15) correct
34%
(01:13)
wrong
based on 128
sessions
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31 Mar 2010, 16:09
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Is x > y ?
(1) (x - y)(x - y) > 0 --> \((x-y)^2>0\) --> the only thing we can deduce from this is that \(x\neq{y}\). (square of a number is always more than or equal to zero. Square of a number equals to zero when the number itself is a zero. Now, as we are given that the square of a number is strictly more than zero thus this number doesn't equal to zero). Not sufficient.
(2) y > 0. Clearly not sufficient.
(1)+(2) All we know is that \(x\neq{y}\) and \(y>0\) --> insufficient to say whether \(x>y\). Not sufficient.
Answer: E.
(1) (x - y)(x - y) > 0 --> \((x-y)^2>0\) --> the only thing we can deduce from this is that \(x\neq{y}\). (square of a number is always more than or equal to zero. Square of a number equals to zero when the number itself is a zero. Now, as we are given that the square of a number is strictly more than zero thus this number doesn't equal to zero). Not sufficient.
(2) y > 0. Clearly not sufficient.
(1)+(2) All we know is that \(x\neq{y}\) and \(y>0\) --> insufficient to say whether \(x>y\). Not sufficient.
Answer: E.
31 Mar 2010, 16:23
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Bunuel
sang5650
Is x > y ?
(I) (x - y)(x - y) > 0
(II) y > 0
(I) (x - y)(x - y) > 0
(II) y > 0
(1) \((x-y)^2>0\) --> the only thing we can deduce from this is that \(x\neq{y}\). Not sufficient.
(2) \(y>0\). Clearly not sufficient.
(1)+(2) \(x\neq{y}\) and \(y>0\). Not sufficient.
Answer: E.
Hi,
My questions is cant we simplify the eq as:
(x-y)^2 > 0
=> (x-y) > 0
=> x > y
Is this method wrong?
