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Is x > y? (1) (x+y)/2 > 0 (2) (y-x)(y+x) > 0

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Is x > y? (1) (x+y)/2 > 0 (2) (y-x)(y+x) > 0 [#permalink] New post 02 May 2008, 15:38
Is x > y?
(1) (x+y)/2 > 0
(2) (y-x)(y+x) > 0
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Re: inequality [#permalink] New post 02 May 2008, 21:14
fresinha12 wrote:
Is x > y?
(1) (x+y)/2 > 0
(2) (y-x)(y+x) > 0


1. (x+y) / 2 > 0
x+y > 0. so nsf.

2. (y-x)(y+x) > 0
y^2 - x^2 > 0
y^2 > x^2. nsf.

E. togather also nsf...
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Re: inequality [#permalink] New post 03 May 2008, 08:49
one of u is correct..

should i divluge the answer???
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Re: inequality [#permalink] New post 03 May 2008, 10:11
I think its E as well

Stat 1 x+y>0
x>-y Insuff
x=1, Y=2
X=2, Y=1

Stat 2

(y-x)(y+x)>0
y>x or y>-x
Insufficient

Together stat 1 and 2 states if y>-x and x>-y but we dont know if x and y are positive or negative so Insufficient
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Re: inequality [#permalink] New post 03 May 2008, 10:55
1. (x+y) / 2 > 0
x+y > 0. so nsf.

2. (y-x)(y+x) > 0
y^2 - x^2 > 0
y^2 > x^2. nsf.

Yes but when you take them together this is what you get :
x+y > 0
(y-x) ( y+x ) > 0

This implies that y-x > 0 ( because the other choices - equal to 0 and less than zero will violate the equation 2 )
Therefore y>x
So its C.
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Re: inequality [#permalink] New post 03 May 2008, 14:07
fresinha12 wrote:
Is x > y?
(1) (x+y)/2 > 0
(2) (y-x)(y+x) > 0


OA is C

1)x+y/2 >0

x+y>0...y>-x..insuff

2) is also insuff..

together..

(y-x)>0 (y+x)>0 if x+y>0.. if x+y>0 then y+x must be >0..

if Y+x>0 then y-x has to be >0..if y-x >0 and y+x>0..then y>x..

sufficient
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Re: inequality [#permalink] New post 03 May 2008, 18:21
I think ans. is C.

Reasoning -

1. (x+y) / 2 > 0
=> x+y>0
With e.g.
Above condn is satisifed for the following (x,y) pairs: (2,-1), (1,2), (-1,2)
So, in one case x>y and in the other two, y>x
Hence, insufficient.

2.(y-x)(y+x) > 0
With e.g.
Above can be re-written as y^2>x^2.
There can be 4 pairs of (x,y) which will satisfy this condn, viz.
(-1,-2), (1,2), (-1,2), (1,-2)
Hence, insufficent

Combining both 1 & 2, we get 2 choices for (x,y) : (1,2) & (-1,2). In both cases y>x.
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Re: inequality [#permalink] New post 03 May 2008, 23:13
I used co-ordinate axis to solve this.
To look at my approach, please have a look at the similar problem, where I have explained my method with a diagram.
http://www.gmatclub.com/forum/7-t63218
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Re: inequality [#permalink] New post 05 May 2008, 15:57
getafixdruid wrote:
1. (x+y) / 2 > 0
x+y > 0. so nsf.

2. (y-x)(y+x) > 0
y^2 - x^2 > 0
y^2 > x^2. nsf.

Yes but when you take them together this is what you get :
x+y > 0
(y-x) ( y+x ) > 0

This implies that y-x > 0 ( because the other choices - equal to 0 and less than zero will violate the equation 2 )
Therefore y>x
So its C.



good one. gotta C...

thanx..............
Re: inequality   [#permalink] 05 May 2008, 15:57
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