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Is x > y? (1) (x+y)/2 > 0 (2) (yx)(y+x) > 0 [#permalink]
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02 May 2008, 16:38
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Is x > y? (1) (x+y)/2 > 0 (2) (yx)(y+x) > 0



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Re: inequality [#permalink]
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02 May 2008, 22:14
fresinha12 wrote: Is x > y? (1) (x+y)/2 > 0 (2) (yx)(y+x) > 0 1. (x+y) / 2 > 0 x+y > 0. so nsf. 2. (yx)(y+x) > 0 y^2  x^2 > 0 y^2 > x^2. nsf. E. togather also nsf...



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Re: inequality [#permalink]
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03 May 2008, 09:49
one of u is correct..
should i divluge the answer???



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Re: inequality [#permalink]
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03 May 2008, 11:11
I think its E as well
Stat 1 x+y>0 x>y Insuff x=1, Y=2 X=2, Y=1
Stat 2
(yx)(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]
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03 May 2008, 11:55
1. (x+y) / 2 > 0 x+y > 0. so nsf.
2. (yx)(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 (yx) ( y+x ) > 0
This implies that yx > 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]
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03 May 2008, 15:07
fresinha12 wrote: Is x > y? (1) (x+y)/2 > 0 (2) (yx)(y+x) > 0 OA is C 1)x+y/2 >0 x+y>0...y>x..insuff 2) is also insuff.. together.. (yx)>0 (y+x)>0 if x+y>0.. if x+y>0 then y+x must be >0.. if Y+x>0 then yx has to be >0..if yx >0 and y+x>0..then y>x.. sufficient



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Re: inequality [#permalink]
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03 May 2008, 19:21
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.(yx)(y+x) > 0 With e.g. Above can be rewritten 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]
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04 May 2008, 00:13
I used coordinate 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/7t63218



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Re: inequality [#permalink]
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05 May 2008, 16:57
getafixdruid wrote: 1. (x+y) / 2 > 0 x+y > 0. so nsf.
2. (yx)(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 (yx) ( y+x ) > 0
This implies that yx > 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..............










