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# MGMAT PS

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Director
Joined: 29 Aug 2005
Posts: 808

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27 Nov 2008, 13:42
1
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If$$/x/-/y/=/x+y/$$ and xy does not equal 0, which of the following must be true?

a) x-y>0
b) x-y0
d)xy>0
e)xy<0

I would appreciate some algebraic solutions to it. Cheers!

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Current Student
Joined: 28 Dec 2004
Posts: 3280
Location: New York City
Schools: Wharton'11 HBS'12

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27 Nov 2008, 16:01
1
If |x|-|y|=|x+y| and xy does not equal 0, which of the following must be true?

a) x-y>0
b) x-y<0
c)x+y>0
d)xy>0
e)xy<0

u shud think in terms of distance on a number line says that distance between x and y is the same x+y

we know that |x| >|y|

pick any example x=-3 and y= 2

3-2=1 =|-3+2|

if x=3 y=-2

3-2=1=|3-2| works..

so you see xy are always <0

E it is
Director
Joined: 14 Aug 2007
Posts: 692

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27 Nov 2008, 18:33
agree with fresinha12.

just to nitpick:
>>we know that |x| >|y|

This is because |x|-|y| > modulus , so |x|-|y| must be greater than 0. i.e |x| >|y|
Manager
Joined: 08 Aug 2008
Posts: 226

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27 Nov 2008, 23:24
for the equation to hold, x must be +ve and y -ve.
hence x*y=-ve<0.
E.
VP
Joined: 17 Jun 2008
Posts: 1474

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28 Nov 2008, 04:41
1
botirvoy wrote:
If$$/x/-/y/=/x+y/$$ and xy does not equal 0, which of the following must be true?

a) x-y>0
b) x-y<0
c)x+y>0
d)xy>0
e)xy<0

I would appreciate some algebraic solutions to it. Cheers!

Algebraic solution:
Square both sides:
Thus, x^2 + y^2 - 2|x||y| = x^2 + y^2 + 2xy
or, 2|x||y| + 2xy = 0.
Since 2|x||y| is always positive, the above equation is possible only when 2xy < 0 or, xy < 0.
Director
Joined: 29 Aug 2005
Posts: 808

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30 Nov 2008, 11:08
scthakur wrote:
Algebraic solution:
Square both sides:
Thus, x^2 + y^2 - 2|x||y| = x^2 + y^2 + 2xy
or, 2|x||y| + 2xy = 0.
Since 2|x||y| is always positive, the above equation is possible only when 2xy < 0 or, xy < 0.

I sometimes fear that by squaring up I might lose certain piece of info, but it seems to work in this case.
OA is indeed E.

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: MGMAT PS &nbs [#permalink] 30 Nov 2008, 11:08
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