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Re: If x # y = –1, which of the following could be true? [#permalink]

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14 Sep 2011, 14:16

Time: 1:00 min.

Approach - Since, x # y = –1 => x#y= (x−y)^2, if x>y - This equation is not possible becasue any square will yield positive value. => x#y= x+y/4, if x≤y - This equation is only possible and as it says - x≤y, Hence - D.

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Re: If x # y = –1, which of the following could be true? [#permalink]

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09 Aug 2017, 00:14

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Re: If x # y = –1, which of the following could be true? [#permalink]

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17 Aug 2017, 22:58

x#y = (x-y)^2 --> this rules out a -ve value (and option which says x > y; becz square can't be -ve) so x#y = -1; we have to consider next equation --which says x < equal to y

if x+y/4= -1 --> x+y = -4 => either x = y (-2 and -2) or x less than y (y is 0; x is -4) Thus I and III

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