IMO it's: B - 2 alone is sufficient.blaster wrote:Is x > y?
(1) x^2 > y
(2) x - |y| > 0
In 1, x & y can be any value. ex. x= -2 and y = 1. x^2 > 1 but x > y is not true. Insufficient.
In 2, x -|y| > 0 is sufficient.
IMO it's: B - 2 alone is sufficient.blaster wrote:Is x > y?
(1) x^2 > y
(2) x - |y| > 0
Because absolute value is always positive or zero:kvcpk wrote:Hi Patrick,Patrick_GMATFix wrote:Hello blaster,
-->On the other hand, if y is negative, then x >y (since x is positive)
-Patrick
How do we know that x is positive?
Got it now.. thanks!!Testluv wrote:Because absolute value is always positive or zero:kvcpk wrote:Hi Patrick,Patrick_GMATFix wrote:Hello blaster,
-->On the other hand, if y is negative, then x >y (since x is positive)
-Patrick
How do we know that x is positive?
x>|y|
x>pos or zero
If x is greater than pos or zero, then x is definitely positive.
Suppose |z| = w
We know the left hand side must be positive or zero. We know the left hand side must equal the right hand side. Thus, w is positive or zero.