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# If != , is / > 1? 1> >0 2> <0

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Manager
Joined: 21 Feb 2007
Posts: 80
If != , is / > 1? 1> >0 2> <0 [#permalink]

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02 Nov 2007, 22:17
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If [x] != [-y] , is [x-y]/[x+y] > 1?

1>[x]>0

2>[y]<0
VP
Joined: 08 Jun 2005
Posts: 1145

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03 Nov 2007, 05:20
This is not the way it was presented on the OG 11 - but here goes:

is (x-y)/(x+y) > 1 ---> when x is different then -y

in order to solve this equation we have to know if (x+y) is positive or negative.

since the possible outcomes are:

x-y > x+y or x-y < x+y

statement 1

x > 0

cannot determine if (x+y) is positive or negative.

insufficient

statement 2

y < 0

cannot determine if (x+y) is positive or negative.

both statements

still insufficient to determine if (x+y) is positive or negative.

Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait

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03 Nov 2007, 05:48
[x-y]/[x+y] > 1?

x-y > x+y ?

only y makes a difference in the equation

So I believe B is the answer
Senior Manager
Joined: 19 Feb 2007
Posts: 325

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03 Nov 2007, 06:00
Ans show be (E)

not y but x+y makes the difference

If (x+y)<0, then
x-y < x+y
Manager
Joined: 21 Feb 2007
Posts: 80

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03 Nov 2007, 14:04
But I am getting C???

If we consider (x-y)>(x+y)
From 1 & 2 and plug in values whatever values we take expression becomes true.
VP
Joined: 08 Jun 2005
Posts: 1145

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03 Nov 2007, 14:12
abiswas wrote:
But I am getting C???

If we consider (x-y)>(x+y)
From 1 & 2 and plug in values whatever values we take expression becomes true.

this is true only when (x+y) > 0 !!!

otherwise you have to flip signs (multiply by a negative number).

Manager
Joined: 21 Feb 2007
Posts: 80

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03 Nov 2007, 16:53
Thanks a ton....KS.
03 Nov 2007, 16:53
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