Inequality

[Kancz44]

Is X> Y?

1) X + 2SqRt(XY) + Y =0
2) X^2 -Y^2=0

OA is A


Thanks!

[aneesh.kg]

Is X> Y?

1) X + 2SqRt(XY) + Y =0
2) X^2 -Y^2=0

OA is A


Thanks!

Statement(1):
((X)^0.5 + (Y)^0.5)^2 = 0
This is possible only when
(X)^0.5 + (Y)^0.5 = 0
This is possible only when
X = Y = 0
Is X > Y?
NO!
SUFFICIENT

Statement(2):
(X - Y)(X + Y) = 0
X = Y or X = - Y
There is no way to determine a relation. With X = -Y, X can be greater, equal or less than Y.
Is X > Y?
DON'T KNOW!
INSUFFICIENT.

[spoiler](A)[/spoiler] is the answer.

P.S. Please use 'spoiler' for OA from next time onwards.

[neelgandham]

Is X> Y?

1) X + 2√(XY) + Y =0

(√X)^2 + 2√(XY) + (√Y)^2 = 0
(√X+√Y)^2 = 0
if square of a number = 0, the number is also equal to 0.
So,√X=-√Y
Since √X,√Y > 0, the only one set of (X,Y) that satisfies the equation √X=-√Y is (0,0). So, X=Y and not >Y.

Statement 1 is sufficient to answer the question.

2) X^2 -Y^2=0

(X-Y)*(X+Y) = 0
Implies X = Y or X = -Y
If Y < 0 and X = -Y, then X>Y
If Y > 0 and X = -Y, then Y>X
and If X = Y then X=Y(lol)
Three different answers!
Statement 2 is insufficient to answer the question.

IMO A