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

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Current Student
Joined: 28 Dec 2004
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Location: New York City
Schools: Wharton'11 HBS'12
Is x > y? (1) (x+y)/2 > 0 (2) (y-x)(y+x) > 0 [#permalink]

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02 May 2008, 16:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 322 [0], given: 2

Director
Joined: 03 May 2007
Posts: 867

Kudos [?]: 271 [0], given: 7

Schools: University of Chicago, Wharton School

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02 May 2008, 22:14
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...

Kudos [?]: 271 [0], given: 7

Current Student
Joined: 28 Dec 2004
Posts: 3345

Kudos [?]: 322 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

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03 May 2008, 09:49
one of u is correct..

should i divluge the answer???

Kudos [?]: 322 [0], given: 2

Senior Manager
Joined: 20 Feb 2008
Posts: 295

Kudos [?]: 51 [0], given: 0

Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross

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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

(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

Kudos [?]: 51 [0], given: 0

Manager
Joined: 22 Dec 2007
Posts: 145

Kudos [?]: 10 [0], given: 0

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03 May 2008, 11:55
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.

Kudos [?]: 10 [0], given: 0

Current Student
Joined: 28 Dec 2004
Posts: 3345

Kudos [?]: 322 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

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03 May 2008, 15:07
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

Kudos [?]: 322 [0], given: 2

Current Student
Joined: 25 Nov 2007
Posts: 31

Kudos [?]: 12 [0], given: 65

Location: India
Schools: ISB '17 (A), IIMA (A)

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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.(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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Intern
Joined: 26 Apr 2008
Posts: 49

Kudos [?]: 3 [0], given: 0

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04 May 2008, 00:13
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

Kudos [?]: 3 [0], given: 0

Director
Joined: 03 May 2007
Posts: 867

Kudos [?]: 271 [0], given: 7

Schools: University of Chicago, Wharton School

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05 May 2008, 16:57
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..............

Kudos [?]: 271 [0], given: 7

Re: inequality   [#permalink] 05 May 2008, 16:57
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# Is x > y? (1) (x+y)/2 > 0 (2) (y-x)(y+x) > 0

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