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# 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]

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12 Dec 2012, 06:46
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36% (01:03) correct 64% (01:33) wrong based on 292 sessions

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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.
Edited OA
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Joined: 02 Sep 2009
Posts: 44319
Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

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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.

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

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12 Dec 2012, 07:01
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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

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]

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14 Dec 2012, 03:12
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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]

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23 May 2013, 23:06
1
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Bunuel wrote:
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.

Hi Bunel,

Can you please explain where am I going wrong:

(x^2+y^2) = x^2+2xy+y^2 = 9a..........(1)

x^2+y^2 >= 2xy ..........(2)

Substitute equation 2 in 1

Thus, (x^2+y^2+x^2+y^2) = 2(x^2+y^2) >= 9a

2(x^2+y^2) >= 9a

Finally, (x^2+y^2) >= 4.5a. Sufficient

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

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Posts: 44319
Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

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24 May 2013, 01:03
Genfi wrote:
Bunuel wrote:
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.

Hi Bunel,

Can you please explain where am I going wrong:

(x^2+y^2) = x^2+2xy+y^2 = 9a..........(1)

x^2+y^2 >= 2xy ..........(2)

Substitute equation 2 in 1

Thus, (x^2+y^2+x^2+y^2) = 2(x^2+y^2) >= 9a

2(x^2+y^2) >= 9a

Finally, (x^2+y^2) >= 4.5a. Sufficient

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

$$x^2+y^2\geq{4.5a}$$ does NOT necessarily mean that $$x^2 + y^2 > 4a$$. Consider x=y=a=0, in this case $$x^2 + y^2= 4a$$.

Hope it's clear.
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Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

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28 May 2013, 15:01
Man i actually made a quick questimate and could tell within 15seconds that it would not work out. so answer I got was E. I have realised a lot of the gmat questions have the same concept and My question is though is this s dangerous way to appraoch the gmat exam
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Joined: 06 Aug 2013
Posts: 89
Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

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23 Oct 2014, 02:08
Bunuel wrote:
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.

Hi bunuel,
a clarification;
in inequalities that have multiple variables as is the case with this question and without any clear information about the variables i.e if they are negative, non-negative etc etc, the options are always insufficient, right?
the reason i am asking this question is because i approached this question the same way and got E.
nothing is mentioned about the variables and most importantly there is more than 1 variable. I mean the values for these variables could be anything. nothing is explicitly mentioned about them.
Math Expert
Joined: 02 Sep 2009
Posts: 44319
Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

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24 Oct 2014, 03:00
arnabs wrote:
Bunuel wrote:
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.

Hi bunuel,
a clarification;
in inequalities that have multiple variables as is the case with this question and without any clear information about the variables i.e if they are negative, non-negative etc etc, the options are always insufficient, right?
the reason i am asking this question is because i approached this question the same way and got E.
nothing is mentioned about the variables and most importantly there is more than 1 variable. I mean the values for these variables could be anything. nothing is explicitly mentioned about them.

No, that's not correct. We do have some information about the variables: we know that a is non-negative, (x + y)^2 = 9a and (x - y)^2 = a. Yes, this info is not enough to answer whether x^2 + y^2 > 4a but if, for example, the question were whether $$x^2 + y^2 \geq{4a}$$, then the answer would be C, not E.
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Re: If a is non-negative, is x^2 + y^2 > 4a ? [#permalink]

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29 Nov 2017, 07:53
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Re: If a is non-negative, is x^2 + y^2 > 4a ?   [#permalink] 29 Nov 2017, 07:53
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