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

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Question Stats: 31% (02:08) correct 69% (02:09) wrong based on 270 sessions

### HideShow timer Statistics If a is non-negative, is x^2 + y^2 > 4a ?

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

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Originally posted by Marcab on 12 Dec 2012, 06:46.
Last edited by walker on 10 May 2013, 09:26, edited 3 times in total.
Edited OA
Math Expert V
Joined: 02 Sep 2009
Posts: 56275
Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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13
3
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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2
1
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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1
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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2
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.

Math Expert V
Joined: 02 Sep 2009
Posts: 56275
Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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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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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
Manager  B
Joined: 06 Aug 2013
Posts: 67
Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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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 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 V
Joined: 02 Sep 2009
Posts: 56275
Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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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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Hi Bunuel, thank you for the detailed explanation.
Another question,
If instead of a being non-negative, it was mentioned that 'a is positive' would the answer be A or C?
I may be wrong, but I suspected A similar to a deduction above i.e x^2+y^2 >= 4.5a
Intern  B
Joined: 06 Jun 2017
Posts: 11
Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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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.

Would it be "C" , if the statement says a is a positive integer ?
Math Expert V
Joined: 02 Sep 2009
Posts: 56275
Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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HariharanIyeer0 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.

Would it be "C" , if the statement says a is a positive integer ?

Yes, because in this case 5a will always be greater than 4a.
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If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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Hi @Bunnel,
Despite going through these comments I still couldn't quite understand the solution at hand.
This is how I solved it, could you please tell me where I went wrong?

Given,
x^2+y^2 > 4a

So, we could write it as
(x+y)^2 - 2xy > 4a
OR
(x-y)^2 + 2xy > 4a

St 1: (x+y)^2 = 9a
So,
9a - 2xy > 4a
-2xy > -5a
2xy > 5a

After plugging in values into the eqn. we get,
(x+y)^2 - 2xy > 4a
9a-6a >! 4a Sufficient.

Similarly I solved St 2 and got 2xy > 3a which is Insufficient.

Hence, I chose A.

Thanks, If a is non-negative, is x^2 + y^2 > 4a ?   [#permalink] 22 Apr 2019, 03:18
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