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If x not equal to -y is x-y/x+y > 1? 1. x > 0 2. y

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Senior Manager
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If x not equal to -y is x-y/x+y > 1? 1. x > 0 2. y [#permalink] New post 14 Jan 2007, 08:47
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A
B
C
D
E

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If x not equal to -y is x-y/x+y > 1?

1. x > 0
2. y < 0
SVP
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 [#permalink] New post 14 Jan 2007, 08:56
(E) for me :)

x-y/x+y > 1 ?
<=> x-y/x+y - 1 > 0 ?
<=> (x-y - (x+y)) / (x+y) > 0
<=> -2*y / (x+y) > 0
<=> y / (x+y) < 0 ?

Sign(y) != Sign(x+y) ?

Stat 1
x > 0. No information about y.

To be sure :
o If y = x = 1, then y/(x+y) = 1/2 > 0
o If y = -2 and x = 3, then y/(x+y) = -2/1 < 0

INSUFF.

Stat 2
y < 0. No information about x.

To be sure :
o If y = x = -1, then y/(x+y) = 1/2 > 0
o If y = -2 and x = 3, then y/(x+y) = -2/1 < 0

INSUFF.

Both (1) & (2)
We need a relationship between x and y.

To be sure :
o If y = -2 and x = 3, then y/(x+y) = -2/1 < 0
o If y = -2 and x = 1, then y/(x+y) = -2/-1 > 0

INSUFF.
Senior Manager
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 [#permalink] New post 14 Jan 2007, 09:14
If x not equal to -y is x-y/x+y > 1?

1. x > 0
2. y < 0

Fig

The way you solved the problem was to simplify the equation by subtracting the equation x-y/x+y - 1> 0.

The way I solved it was use statement 1 in the equation and since I did not know the value of y I said statement 1 was insufficient

Similarly from statement 2 since x was unknown I said statement 2 is insufficient. Combining both the statements I just took 2 values and tried putting it in the equation and found that the 2 values were giving me diff results. Hence I concluded it to be E.

Is my approach correct?
Director
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 [#permalink] New post 14 Jan 2007, 09:23
Fig wrote:
(E) for me :)

x-y/x+y > 1 ?
<=> x-y/x+y - 1 > 0 ?
<=> (x-y - (x+y)) / (x+y) > 0
<=> -2*y / (x+y) > 0
<=> y / (x+y) < 0 ?

Sign(y) != Sign(x+y) ?

Stat 1
x > 0. No information about y.

To be sure :
o If y = x = 1, then y/(x+y) = 1/2 > 0
o If y = -2 and x = 3, then y/(x+y) = -2/1 < 0

INSUFF.

Stat 2
y < 0. No information about x.

To be sure :
o If y = x = -1, then y/(x+y) = 1/2 > 0
o If y = -2 and x = 3, then y/(x+y) = -2/1 < 0

INSUFF.

Both (1) & (2)
We need a relationship between x and y.

To be sure :
o If y = -2 and x = 3, then y/(x+y) = -2/1 < 0
o If y = -2 and x = 1, then y/(x+y) = -2/-1 > 0

INSUFF.


I guess the question has to be understood like this:

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

No? :lol:
VP
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Location: Bangalore
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Re: OG 139 DS [#permalink] New post 14 Jan 2007, 09:33
axl_oz wrote:
If x not equal to -y is x-y/x+y > 1?

1. x > 0
2. y < 0


Question can be written as :
=>Is x-y > x+y?
=>Is -y>+y?
=>Is -2y>0?
=> Is y>0?

1. Insuff
2. Suff

My answer is B.
Director
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Re: OG 139 DS [#permalink] New post 14 Jan 2007, 09:38
kripalkavi wrote:
axl_oz wrote:
If x not equal to -y is x-y/x+y > 1?

1. x > 0
2. y < 0


Question can be written as :
=>Is x-y > x+y?
=>Is -y>+y?
=>Is -2y>0?
=> Is y>0?

1. Insuff
2. Suff


My answer is B.


I think we can't cross multiply in inequalities. So x-y/x+y > 1 is not same as x-y > x+y.
SVP
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 [#permalink] New post 14 Jan 2007, 09:50
Yes, it is :) Your approach is perfectly fine :)

I always prefer to compare signs of expression : it's generally easier to conclude on :)

axl_oz wrote:
If x not equal to -y is x-y/x+y > 1?

1. x > 0
2. y < 0

Fig

The way you solved the problem was to simplify the equation by subtracting the equation x-y/x+y - 1> 0.

The way I solved it was use statement 1 in the equation and since I did not know the value of y I said statement 1 was insufficient

Similarly from statement 2 since x was unknown I said statement 2 is insufficient. Combining both the statements I just took 2 values and tried putting it in the equation and found that the 2 values were giving me diff results. Hence I concluded it to be E.

Is my approach correct?
SVP
SVP
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Joined: 01 May 2006
Posts: 1813
Followers: 8

Kudos [?]: 92 [0], given: 0

 [#permalink] New post 14 Jan 2007, 09:52
Yes :)..... it's the x != -y that helps us to conclude so :)

aurobindo wrote:

I guess the question has to be understood like this:

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

No? :lol:
Senior Manager
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Re: OG 139 DS [#permalink] New post 14 Jan 2007, 16:27
(1) (plug in #s) Let x = 1: (1-y)/(1+y) could be > or < 1 depending on y. Insuff => B, C or E.

(2) (plug in #s) Let y = -1: (x+1)/(x-1). Like in (1), this expression can be > or < 1 depending on x. Insuff => C or E.

(1&2) (plug in #s) Let (x,y) = (10,-1): 11/9 > 1. Cool so far. Now, Let (x,y) = (1,-10): 11/-9 < 1 => Insuff.

E.
Re: OG 139 DS   [#permalink] 14 Jan 2007, 16:27
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