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1) insuff. 4 cases. a)y>0 and y<x answer is yes b)y>0 and y>x answer is yes c)y<0 and absolute value y>x answer is yes d)y<0 and absolute value y<x answer is no 2)insuff 4 cases a and b is like the above where is x>0 and answer is yes c)x<0 and y<x answer is yes d)x<0 and y>x answer is no

together x>0 and y<0 insuff depends on the comparison between the absolute value of y and x.

Re: Is (x-y)/ (x+y) < 1 1) x > 0 2) y < 0 [#permalink]
09 Oct 2013, 07:53

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Re: If x#-y is (x-y)/(x+y)>1? [#permalink]
09 Oct 2013, 08:01

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Expert's post

If x\neq{-y} is \frac{x-y}{x+y}>1?

Is \frac{x-y}{x+y}>1? --> Is 0>1-\frac{x-y}{x+y}? --> Is 0>\frac{x+y-x+y}{x+y}? --> Is 0>\frac{2y}{x+y}?

(1) x>0 --> Not sufficient.

(2) y<0 --> Not sufficient.

(1)+(2) x>0 and y<0 --> numerator (y) is negative, but we cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.