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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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New post Updated on: 10 May 2013, 09:26
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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

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

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New post 12 Dec 2012, 06:55
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.

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

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New post 12 Dec 2012, 07:01
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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New post 14 Dec 2012, 03:12
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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New post 23 May 2013, 23:06
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.

Answer: E.



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.

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

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New post 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.

Answer: E.



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.

Answer: A


\(x^2+y^2\geq{4.5a}\) 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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New post 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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Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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New post 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.

Answer: E.


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

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New post 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.

Answer: E.


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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New post 27 Sep 2018, 01:14
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
What are your thoughts?
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Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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New post 02 Nov 2018, 15:04
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.

Answer: E.



Would it be "C" , if the statement says a is a positive integer ?
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Re: If a is non-negative, is x^2 + y^2 > 4a ?  [#permalink]

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New post 03 Nov 2018, 01:38
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.

Answer: E.



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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New post 22 Apr 2019, 03:18
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.
Please help me understand this problem.

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