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# if x is not equal to -y is x-y/x+y > 1 1. x>0 2.

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SVP
Joined: 28 Dec 2005
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if x is not equal to -y is x-y/x+y > 1 1. x>0 2. [#permalink]  19 Mar 2008, 18:19
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if x is not equal to -y

is x-y/x+y > 1

1. x>0
2. y<0
Manager
Joined: 20 Aug 2007
Posts: 65
Followers: 1

Kudos [?]: 9 [0], given: 0

Re: DS OG [#permalink]  19 Mar 2008, 20:01
pmenon wrote:
if x is not equal to -y

is x-y/x+y > 1

1. x>0
2. y<0

Fact 1
if X>0, no conclusions can be made
eg. X = 5 Y = 1 then <1
if X = 5 and Y = -1 then >1

Fact 2
if y<0, no conclusions can be made
eg. Y=-1 X=-5, then <1
if Y=-1 X = 5, then >1

Combined facts: no conclusions can be made
if X = 5 Y = -1, then >1
if X = 5 and T = -10, then <1

CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 405

Kudos [?]: 2123 [0], given: 359

Re: DS OG [#permalink]  19 Mar 2008, 20:58
Expert's post
pmenon wrote:
if x is not equal to -y

is x-y/x+y > 1

1. x>0
2. y<0

$$\frac{x-y}{x+y}=\frac{x+y-2y}{x+y}=1-\frac{2y}{x+y}>1$$

$$\frac{2y}{x+y}<0$$

Now, we can construct our examples:
1. 2y is always negative (2. y<0)
2. a sign of (x+y) depends on relationship between |x| and |y|. if |x|>|y| then (x+y)>0, otherwise, if |y|<|x| then (x+y)<0
3. x=1, y=-10 and x=10, y=-1
4. Therefore, both conditions are insufficient.
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Director
Joined: 10 Sep 2007
Posts: 949
Followers: 7

Kudos [?]: 204 [0], given: 0

Re: DS OG [#permalink]  19 Mar 2008, 21:04
Given in question that X and -Y are not equal. So X+Y will not be zero. Thereby telling that denominator is not Zero.

We need to find x-y/x+y is greater than 1 or not.

Statement 1:
Tells us x is +ve. But nothing is said about y. Moreover values of x and y are also not mentioned. so question cannot be answered using this statement alone.

Statement 2:
Tells us that y is negative. But nothing is said about x. Moreover values of x and y are also not mentioned. so question cannot be answered using this statement alone.

Combining both statements:
We have x is +ve and y is -ve, which employs x-y (numerator is +ve).
Now coming to denominator x+y. This term can be -ve if |y|>|x|, or this term can be +ve if |x|>|y|.
Since this we do not know what the values of x and y are, so this question cannot be answered.

SVP
Joined: 28 Dec 2005
Posts: 1583
Followers: 2

Kudos [?]: 86 [0], given: 2

Re: DS OG [#permalink]  20 Mar 2008, 03:10
great, you guys are getting the right answers. here is my question. i initially started off rearranging the stem to make the question simpler. my first step was to cross multiply to get:

x-y > x+y

x-y-x-y > 0

-2y > 0

y < 0

so my answer was B. Where did i go wrong ?
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 405

Kudos [?]: 2123 [0], given: 359

Re: DS OG [#permalink]  20 Mar 2008, 03:29
Expert's post
pmenon wrote:
great, you guys are getting the right answers. here is my question. i initially started off rearranging the stem to make the question simpler. my first step was to cross multiply to get:

x-y > x+y

x-y-x-y > 0

-2y > 0

y < 0

so my answer was B. Where did i go wrong ?

Your transition from $$\frac{x-y}{x+y}>1$$ to $$x-y>x+y$$ is wrong.

The correct transition has to be:

$$x-y>x+y$$ for $$x+y>0$$

$$x-y<x+y$$ for $$x+y<0$$

So, You lost important part of the transition
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Re: DS OG   [#permalink] 20 Mar 2008, 03:29
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