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If a is nonnegative, is x^2 + y^2 > 4a ?
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Updated on: 10 May 2013, 09:26
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If a is nonnegative, 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




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Re: If a is nonnegative, is x^2 + y^2 > 4a ?
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12 Dec 2012, 06:55
Marcab wrote: If a is nonnegative, 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 nonnegative, 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^22xy+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 nonnegative, is x^2 + y^2 > 4a ?
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12 Dec 2012, 07:01
If a is nonnegative, 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 nonnegative, is x^2 + y^2 > 4a ?
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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 nonnegative), in that case the answer becomes may be. Therefore the answer is (E).
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Re: If a is nonnegative, is x^2 + y^2 > 4a ?
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23 May 2013, 23:06
Bunuel wrote: Marcab wrote: If a is nonnegative, 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 nonnegative, 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^22xy+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^22xy+y^2=a. Clearly insufficient. Answer: A



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Re: If a is nonnegative, is x^2 + y^2 > 4a ?
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24 May 2013, 01:03
Genfi wrote: Bunuel wrote: Marcab wrote: If a is nonnegative, 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 nonnegative, 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^22xy+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^22xy+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 nonnegative, is x^2 + y^2 > 4a ?
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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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Re: If a is nonnegative, is x^2 + y^2 > 4a ?
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23 Oct 2014, 02:08
Bunuel wrote: Marcab wrote: If a is nonnegative, 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 nonnegative, 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^22xy+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, nonnegative 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 nonnegative, is x^2 + y^2 > 4a ?
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24 Oct 2014, 03:00
arnabs wrote: Bunuel wrote: Marcab wrote: If a is nonnegative, 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 nonnegative, 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^22xy+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, nonnegative 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 nonnegative, (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 nonnegative, is x^2 + y^2 > 4a ?
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27 Sep 2018, 01:14
Hi Bunuel, thank you for the detailed explanation. Another question, If instead of a being nonnegative, 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 nonnegative, is x^2 + y^2 > 4a ?
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02 Nov 2018, 15:04
Bunuel wrote: Marcab wrote: If a is nonnegative, 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 nonnegative, 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^22xy+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 nonnegative, is x^2 + y^2 > 4a ?
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03 Nov 2018, 01:38
HariharanIyeer0 wrote: Bunuel wrote: Marcab wrote: If a is nonnegative, 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 nonnegative, 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^22xy+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 nonnegative, is x^2 + y^2 > 4a ?
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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 (xy)^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 9a6a >! 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, Padmini




If a is nonnegative, is x^2 + y^2 > 4a ?
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22 Apr 2019, 03:18






