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x^2 + y^2 < 6?

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Source: — Data Sufficiency |
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by sumanr84 » Wed Jan 20, 2010 10:11 am
I think C is correct.
Statement 1: (x+y)^2 < 6
x^2+y^2+2xy < 6, Insufficient as it depends on the signs of (x and y)


Statement 2: xy =2 , Insufficient

Taking 1 & 2 together,

using 2 in 1, x^2 +y^2 + 2. 2 < 6

x^2 +y^2 < 2, and definitely < 6 ( answers the Q )
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by GhassanMBA » Wed Jan 20, 2010 11:52 am
I'm getting C as well. Is there anyone else here that is getting E?
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by kewltot » Wed Jan 20, 2010 1:47 pm
^2 + y^2 < 6?

1) (x+y)^2 < 6
2) xy = 2


Option 1 can be written as x^2+y^2+2 x y < 6

Option 2 can be written as y = 2/x

Using 2 into 1

x^2+ (2/x)^2+ 2 x (2/x) < 6

x^2 + (2/x)^2 + 4 < 6

x^2 + (2/x)^2 < 6-4

=> x^2 + (2/x)^2 < 2

=> x^2 + y^2 < 2 (since y = 2/x)

so, we could determine that x^2 + y^2 < 2 but not x^2 + y^2 < 6

OA is C
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by onedayi'll » Wed Jan 20, 2010 3:56 pm
yes C is the answer.
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