41) MOD DS

[ern5231]

Is x^2-y^2> 0?

1) x>|y|
2) |x|>y

[liferocks]

we can re write the question as
IS (x+y)(x-y)>0

From 1 x>|Y|

hence X is positive ..so (x+y) and (x-y) will be both positive...(x+y)(x-y)>0 ..sufficient

From 2
|x|>y


So (x+y)(x-y)>0 ..sufficient
Ans option D

[ballubalraj]

Another way to solve it:

Square of any positive or negative number is always positive. So irrespective of the sign of x and y, as long as x > y (leaving the sign), x^2 - y^2 will always be > 0.

As both the statements say that x > y (leaving the sign part), both statements alone are sufficient.

Answer: D

[November Rain]

Hi,

I agree that statment 1 is sufficient, but i am not sure about 2.

take X = -3 and Y = -4

This means that |x|>y, but, x^2 - y^2 < 0

[student22]

Can't you just simply square both sides of each statement? That way we don't have to worry about sign changes. Or am I oversimplifying it?

[ballubalraj]

I guess November Rain is correct. GMAT always tricks us with simple looking but tricky questions - ineuqalities and modes are probably the best examples.

So the answer should be A.

Any thoughts?

[nakul_anand]

Yep...I think that the answer should be A

November Rain's values - -3 and -4 prove that a^2-b^2 can be less than 0.

[ern5231]

Also does not work out when x=2 and y=-1