x^2 - y ^2 > 0
(x + y)(x-y) > 0

The question asking "x-y > 0".

1. x > | y|
x + y > |y| + y >= 0
x+y is +ve, x^2 - y^ is +ve, therfore x-y must be positive. Sufficient.

2. |x| > y
x + |x| > y + x
y + x < x + |x|
x+|x| = 0 or 2|x|, from this, we can say that x+y can be negative or +ve.


or |x| -x > y -x
|x| -x = 0 when x is +ve
|x -x = -2x when x is -ve
|x|-x's max value is a positive number.
Therfeore, y-x can be negative or positive.



Insufficient.

Well, there are other ways to solve: geometric approach, which I found cumbersome, unless algebraic approach makes you lead nowhere.

Here, i can find a way to determine sign of x+y or x-y algebraically.

heshamelaziry wrote:Q. If x^2 > y^2, is x>y?
1) x > |y|
2) |x| > y


I hate the GMAT. IMO A.

PS: this is categorized as easy.

I think it's C

given x^2 > y^2.. we know that
either x - y > 0 and x +y >0
or x-y<0 and x+y<0

then we look at ind'l statements.
1. x> abs y
therefore x>y or x>-y
which implies x -y >0 or x+y>0
Not Suff


2. abs x -y>0
x-y>0 or x+y<0
Not suff

Taken together.. we find that only x-y>0 and

from the given.. if x-y>0 then x+y>0
therefore x>y or x >-y

Hence C

Please let me know the OA.

KICKGMATASS123 wrote:

heshamelaziry wrote:Q. If x^2 > y^2, is x>y?
1) x > |y|
2) |x| > y


I hate the GMAT. IMO A.

PS: this is categorized as easy.

I think it's C

given x^2 > y^2.. we know that
either x - y > 0 and x +y >0
or x-y<0 and x+y<0

then we look at ind'l statements.
1. x> abs y
therefore x>y or x>-y
which implies x -y >0 or x+y>0
Not Suff

When x + y > 0, given x^2 - y^2 >0, we can conclude that x -y > 0

Thats sufficient on its own.

Can you explain HOW is this possible?????


given x^2 > y^2.. we know that
either x - y > 0 and x +y >0
or x-y<0 and x+y<0

-

brick2009 wrote:Can you explain HOW is this possible?????


given x^2 > y^2.. we know that
either x - y > 0 and x +y >0
or x-y<0 and x+y<0

-

1. x > |y|

look at the case 1: x > y, when y is positive.

look at the case 2: x > -y when y is negative.
x > -y is same as x +y > 0

but we know (x+y)(x-y) > 0
since x+y is positive, from the above, we can conclude that x-y is +ve.

In both cases, x > y is true. Sufficient.

This is not the case in (2).