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# If 0<x<y, what is the value of (x+y)^2 / (x-y)^2 ? 1.

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Joined: 18 Jul 2008
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If 0<x<y, what is the value of (x+y)^2 / (x-y)^2 ? 1. [#permalink]

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12 Aug 2008, 11:29
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If 0<x<y, what is the value of (x+y)^2 / (x-y)^2 ?
1. x^2 + y^2 = 3xy
2. xy = 3
SVP
Joined: 07 Nov 2007
Posts: 1789
Location: New York
Re: one more: gmatperp DS [#permalink]

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12 Aug 2008, 11:35
mba9now wrote:
If 0<x<y, what is the value of (x+y)^2 / (x-y)^2 ?
1. x^2 + y^2 = 3xy
2. xy = 3

(x+y)^2 / (x-y)^2 = ( x^2 + y^2+2xy )/( x^2 + y^2-2xy )

1) suffcieint
( x^2 + y^2+2xy )/( x^2 + y^2-2xy )
= 5xy/xy=5
2) ( x^2 + y^2+2xy )/( x^2 + y^2-2xy ) = (x^2 + y^2+6)/( x^2 + y^2-6)
two variable and only one equation xy=3 and we need to find values both variables.
Not suffcieint

A.

What is OA
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Re: one more: gmatperp DS [#permalink]

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12 Aug 2008, 11:49
A

#1

$$\frac{(x+y)^2}{(x-y)^2} =$$

$$\frac{(x+y)(x+y)}{(x-y)(x-y)}=$$

$$\frac{x^2+2xy+y^2}{x^2-2xy+y^2}$$

$$\frac{x^2+y^2+2xy}{x^2+y^2-2xy}$$

Now substitute $$3xy$$ for $$x^2 + y^2$$ because $$x^2 + y^2 = 3xy$$

$$\frac{3xy+2xy}{3xy-2xy}$$

$$\frac{5xy}{xy} = 5$$ SUFFICIENT

$$\frac{x^2+y^2+2xy}{x^2+y^2-2xy}$$

Now substitute 3 for xy:

$$\frac{x^2+y^2+2(3)}{x^2+y^2-2(3)}$$

$$\frac{x^2+y^2+6}{x^2+y^2-6}$$

It can't be simplieifed any further and we don't know values of x or y. INSUFFICIENT.

mba9now wrote:
If 0<x<y, what is the value of (x+y)^2 / (x-y)^2 ?
1. x^2 + y^2 = 3xy
2. xy = 3

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Joined: 23 Apr 2008
Posts: 85
Re: one more: gmatperp DS [#permalink]

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13 Aug 2008, 22:30
mba9now wrote:
If 0<x<y, what is the value of (x+y)^2 / (x-y)^2 ?
1. x^2 + y^2 = 3xy
2. xy = 3

A.
(x+y)^2=x^2+y^2+2xy
(x-y)^2=x^2+y^2-2xy

on simplification..we are left with 5xy/xy => 5

A IS SUFFICIENT

B.WE CANT FIND THE VALUE WITH THIS INFORMATION

Re: one more: gmatperp DS   [#permalink] 13 Aug 2008, 22:30
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