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# What is the value of x^2 + y^2 ? (1) x^2+ y^2 = 2xy + 1 (2)

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What is the value of x^2 + y^2 ? (1) x^2+ y^2 = 2xy + 1 (2) [#permalink]

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05 Jun 2011, 09:16
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Question Stats:

50% (00:00) correct 50% (03:00) wrong based on 8 sessions

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What is the value of$$x^2 + y^2$$?
(1) $$x^2+ y^2$$ = 2xy + 1
(2)$$x^2 + y^2$$= 4 - 2xy
Senior Manager
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05 Jun 2011, 09:39
x^2 + y^2 ?

Stmt1: x^2 + y^2 = 2xy+1 Insufficient
Stmt2: x^2+y^2 = 4-2xy Insufficient

Together,
2xy + 1 = 4 - 2xy
4xy=3
xy=3/4
So, x^2+y^2 = 2*3/4 + 1
= 5/2

OA C.
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Manager
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05 Jun 2011, 17:14
OA was E.

But i agree that it should be C. If anyone thinks OA is correct please post your explanation. Thanks!
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05 Jun 2011, 19:44
I too think the answer is C. I did the following to verify the calcualtion:

(1) is insufficient

(x-y)^2 = 1

x-y = -1 or x-y = 1

(2) is insufficient

(x+y)^2 = 4

X+y = 2 or x+y = -2

So we have the following values of x and y

x-y = -1
X+y = 2

x = 1/2, y = 3/2

x-y = 1
X+y = 2

x = 3/2, y = 1/2

x-y = -1
x+y = -2

x = -3/2 y = -1/2

x-y = -1
x+y = -2

x = -3/2 y = -1/2

So value of x and y in each case should be same

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06 Jun 2011, 00:10
a x-y = +|- 1 not sufficient.

b x+y = +|- 2 not sufficient.

a+b solving we get x = +|- 1.5 and +|- 0.5 similarly y = +|- 1.5 and +|- 0.5.

hence x^2 + y^2 = (1.5)^2 + (0.5)^2 = 2.25 + 0.25 = 2.5 (for fixed set of x,y as solutions)

Thus C.
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Re: Binomial DS   [#permalink] 06 Jun 2011, 00:10
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