X is even

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by kvcpk » Fri Jul 02, 2010 10:45 pm
taneja.niks wrote:another one guys

If x is an integer ,is x even?
1. x2-y2=0
2. x2+y2=18
x2-y2=0
let x=2, then y=2 or -2 [x is even]
x=3, then y=3 or -3 [x is odd]

INSUFFICIENT

x2+y2=18
x=3, y=3 or -3 [x is odd]
x=4, y=root(2) [x is even]

INSUFFICIENT

combining both, we get,
2x^2 = 18,
x^2=9
x = 3 or -3
answer is NO.

Pick C

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by Rahul@gurome » Fri Jul 02, 2010 10:49 pm
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).
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