Is x2 + y2 > 6

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by ayushiiitm » Fri Jun 25, 2010 5:46 pm
stem 1 says (x+y)^2>6

i.e x^2+y^2+2xy>6

you may be tempted here to conclude that then (x+y)^2<6

think of the case now like this
x^2+y^2+2xy>6

then may be x^2+y^2 was equal to 4 and 2xy was equal to 3 so x^2+y^2 <6, even though x^2+y^2+2xy>6
or may be x^2+y^2 was equal to 8 and 2xy was equal to 1 so x^2+y^2 >6 even though x^2+y^2+2xy>6

insufficient

stem 2 is not sufficient as we see


using both stem 1 and 2

x^2+y^2+2xy>6....putting value of xy
x^2+y^2>4

so x^2+y^2 could be equal to 5

or x^2+y^2 could be equal to 10

so insufficient
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by naki009 » Fri Jun 25, 2010 7:34 pm
ayushiiitm wrote:stem 1 says (x+y)^2>6

i.e x^2+y^2+2xy>6

you may be tempted here to conclude that then (x+y)^2<6

think of the case now like this
x^2+y^2+2xy>6

then may be x^2+y^2 was equal to 4 and 2xy was equal to 3 so x^2+y^2 <6, even though x^2+y^2+2xy>6
or may be x^2+y^2 was equal to 8 and 2xy was equal to 1 so x^2+y^2 >6 even though x^2+y^2+2xy>6

insufficient

stem 2 is not sufficient as we see


using both stem 1 and 2

x^2+y^2+2xy>6....putting value of xy
x^2+y^2>4

so x^2+y^2 could be equal to 5

or x^2+y^2 could be equal to 10

so insufficient
Thank you.. thats a really nice explanation

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by Haaress » Sat Jun 26, 2010 2:52 pm
Good question. Thanks for posting it.

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by swiftwolff » Sat Feb 04, 2012 7:12 pm
I find (2)xy=2 sufficient somehow... because there are only limited possibilities for the product of x and y, It could be x=2 y=1, x=1 y=2, x=-2 y=-1, x=-1 y=-2, x=square root of 2 y=square root of 2 (also negative square root of 2). When square all of the set, all of the answer sets will add up to be less than 6. Can anyone comment on this?

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by swiftwolff » Sat Feb 04, 2012 7:14 pm
Never mind, x could be 6 and y could be 1/3.... dumb me... sorry

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by Anurag@Gurome » Sat Feb 04, 2012 8:47 pm
naki009 wrote:Is x^2 + y^2 > 6?
(1) (x + y)^2 > 6
(2) xy = 2

OA : E

I can anyone please explain the answer for this ?
(1) (x + y)² > 6 implies x² + y² + 2xy > 6
If x² + y² = 5 and 2xy = 3, then x² + y² < 6
If x² + y² = 10 and 2xy = 3, then x² + y² > 6
No definite answer; NOT sufficient.

(2) xy = 2
If x = y = √2, then x² + y² = 4 < 6
If x = 10, y = 1/5, then x² + y² = 100 + 1/25 > 6
No definite answer; NOT sufficient.

Combining (1) and (2), x² + y² + 2xy > 6
x² + y² + 2(2) > 6
x² + y² + 4 > 6
x² + y² > 2
If x² + y² = 5, then x² + y² < 6
If x² + y² = 10 , then x² + y² > 6
No definite answer; NOT sufficient.

The correct answer is E.
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