If a is non-negative, is x^2 + y^2 > 4a ?

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If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

12 Dec 2012, 06:46

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If a is non-negative, is x^2 + y^2 > 4a ?

(1) (x + y)^2 = 9a
(2) (x - y)^2 = a

[Reveal] Spoiler: OA

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Last edited by walker on 10 May 2013, 09:26, edited 3 times in total.

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Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

12 Dec 2012, 06:55

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Marcab wrote:

If a is non-negative, is x^2 + y^2 > 4a ?

(1) (x + y)^2 = 9a
(2) (x - y)^2 = a

Source: Jamboree
I am not convinced with the OA.

If a is non-negative, is x^2 + y^2 > 4a ?

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x, y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.
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Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

12 Dec 2012, 07:01

If a is non-negative, is x^2 + y^2 > 4a ?
(1) (x + y)^2 = 9a
(2) (x - y)^2 = a

I am getting E
A is non negative, so a can be zero or positive.
x=0, y =0, a = 0...satisfies both 1) and 2) but not x^2 + y^2>4a

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Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

14 Dec 2012, 03:12

Ans:

both the statements alone don’t give any solution . When we combine them and add we get x^2+y^2=5a which is greater than 4a , but a can be zero( a is non-negative), in that case the answer becomes may be. Therefore the answer is (E).
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Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink] 14 Dec 2012, 03:12

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If a is non-negative, is x^2 + y^2 > 4a ?

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