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Re: Q128 OG 11th Ed [#permalink]
19 Apr 2009, 20:37

pbanavara wrote:

If (x-y)/(x+y) > 1 Then

x-y > x+y => 2y>0 or y>0

From B You have y<0 which answers the question. I haven't looked at the answer but is this correct ?

-pradeep

Nope that's not it. You have to consider (x+y) could be negative. In which case the inequality changes - not completely sure how. Even after I did that I didn't get the answer though

Re: Q139 OG 11th Ed [#permalink]
21 Apr 2009, 14:24

bandit wrote:

Is x-y/x+y>1 ? or we can say is => (x-y/x+y)-1>0 => -2y/x+y >0 => 2y/x+y<0?

St 1: x>0 Not sufficient to tell whether 2y/x+y < 0

St 2: 2) y<0

Not sufficient to tell whether 2y/x+y < 0

Combining also we cannot say 2y/x+y < 0, therefore E

@bandit : Great technique to look at the equation whether it is less than or equal to zero. But, When we combine then, then we need to use the numbers to confirm that the answer is truly E.

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