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If x not equal to -y, is (x-y)/(x+y) > 1

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If x not equal to -y, is (x-y)/(x+y) > 1 [#permalink] New post 19 Apr 2009, 20:19
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If x not equal to -y, is (x-y)/(x+y) > 1 ?

1) x>0
2) y<0

The answer explanation uses "number picking" to solve. Is there another way?

Last edited by thinkblue on 19 Apr 2009, 20:35, edited 1 time in total.
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Re: Q128 OG 11th Ed [#permalink] New post 19 Apr 2009, 20:24
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
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Re: Q128 OG 11th Ed [#permalink] New post 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 :)
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Re: Q139 OG 11th Ed [#permalink] New post 20 Apr 2009, 04:40
so what is the right answer?

I think it is E
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Re: Q139 OG 11th Ed [#permalink] New post 20 Apr 2009, 06:02
even I got E.. but again by number picking.

For S1.. X>0
Case 1: x=4 y=8 Fraction = -1/3 i.e. <1

Case 2: x=4 y=-6 Fraction = -5 <1

Case 3: x=4 y=-2 Fraction = 3 > 1

For S2.. y< 0
Cas2 and Case 3 proves that S2 is not sufficient

Case 1 & Case 2 also proves that S1 and S2 together are also not sufficient.

But will be good to know if this can be done without number picking ... more than individual signs of X and Y, sign of (X+Y) is more vital.

Thanks
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Re: Q139 OG 11th Ed [#permalink] New post 20 Apr 2009, 11:27
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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
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Re: Q139 OG 11th Ed [#permalink] New post 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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Re: Q139 OG 11th Ed [#permalink] New post 22 Apr 2009, 20:22
I could not see how to solve it other than by picking numbers. :|
However, I got E.
Re: Q139 OG 11th Ed   [#permalink] 22 Apr 2009, 20:22
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