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Source: — Data Sufficiency |
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by leonswati » Sun Oct 16, 2011 6:00 pm
Is the answer B? Wats the OA?
Swati
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by neelgandham » Sun Oct 16, 2011 6:38 pm
lenagmat wrote:Is (X^2+Y^2) a square of an integer?

1). x is even, y is odd
2). x and y are odd

1) 3^2 + 4^2 = 5^2
15^2 + 16^2 != integer ^2

So, you cannot answer the question, with this data !

2) x and y are odd

Let us write x and y in the form of 2a+1, 2b+1 were a,b are integers. When you square and add these numbers, we get the below.

4(a^2+b^2+a+b) + 2 = 2*(2(a^2+b^2+a+b)+1)

Let c = (a^2+b^2+a+b)

Then we get the final expression = 2* (2*c +1) = Odd * 2 , which can never be a square of a number!

Hence option B
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by Anurag@Gurome » Sun Oct 16, 2011 8:22 pm
lenagmat wrote:Is (X^2+Y^2) a square of an integer?

1). x is even, y is odd
2). x and y are odd
(1) x is even, y is odd.
If x = -2 and y = 5, then x² + y² = 4 + 25 = 29, which is not the square of an integer.
If x = 4 and y = 3, then x² + y² = 16 + 9 = 25, which is the square of an integer.
We do not get a definite answer; NOT sufficient.

(2) x and y are odd.
If x = 1 and y = 3, then x² + y² = 1 + 9 = 10, which is not the square of an integer.
If x = 3 and y = 5, then x² + y² = 9 + 25 = 34, which is not the square of an integer.
So, x² + y² is not the square of an integer; SUFFICIENT.

The correct answer is B.
Anurag Mairal, Ph.D., MBA
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by lenagmat » Sun Oct 16, 2011 11:27 pm
Thank you. The OA is B.
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