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by kmittal82 » Tue Oct 25, 2011 5:54 am
Let A = x, B = 2x, C = 3x

Take option (D)
3A^2 + B^2 + C^2

= 3x^2 + 4x^2 + 9x^2
= 16x^2

Sqrt of this will be 4x, which is always an integer

Ought to be (D)

OA please?
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by neelgandham » Tue Oct 25, 2011 6:03 am
MBA.Aspirant wrote:confused between C and D
a:b:c = 1:2:3, Substitute a=1,b=2,c=3 in the options and you get 6,14,36,16,55 So, Answer : C or D

Substitute a=2,b=4,c=6, C) 8+64+216 = 288 != square So D is the answer !
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by mankey » Tue Oct 25, 2011 9:30 am
Dont know how to eliminate one of C and D. To me both satisfy the given conditions.

Please help.

Thanks.
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by neelgandham » Tue Oct 25, 2011 10:25 am
mankey wrote:Dont know how to eliminate one of C and D. To me both satisfy the given conditions.

Please help.

Thanks.
Let A = x, B = 2x, C = 3x

Take option (D) 3A^2 + B^2 + C^2 = 3x^2 + 4x^2 + 9x^2 = 16x^2, Sqrt of this will be 4x, which is always an integer


Take option (C) A^3 + B^3 + C^3 = x^3 + 8x^3 + 27x^3 = 36x^3, Sqrt of this will be 6x * Square root(x), is an integer only if x is a perfect square.

Hence, option D. Happy ?
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