Consider (1) alone first.
Given x2-y2 = 0. So (x+y)(x-y) = 0. Or x = -y or x = y.
From this we cannot judge whether x is even or not.
Examples are x= +2 and y = -2, x = +3 and y = 3.
In both case x2 - y2 is 0 but in one case x is even and in other case it is odd.
Or (1) alone is not sufficient.
Consider (2) alone first.
Let x = 2, y = sqrt(14). Here x2+y2 = 4 + 14 = 18. Here x is even.
Let x = 3, y = 3. Here x2+y2 = 9+9 = 18. Here x is odd.
Since we cannot say definitely whether x is even or odd, (2) alone is not sufficient.
Next combine both the statements together and check.
Add x2-y2=0 and x2+y2 = 18.
We have 2*x2 = 18.
Or x2 = 9.
Or x = +3 or -3.
Or we can say x is odd.
So x is not even.
The correct answer is hence (C).
Rahul Lakhani
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Gurome, Inc.
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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