grandh01 wrote:Does x + y = 0?
(1) xy<0
(2) x^2 = y^2

OA is C

Statement 1: xy<0
From this information, there are several possible cases. Here are two.
Case a: x=1, y=-1 in which case x+y equals zero.
Case b: x=2, y=-1 in which case x+y does not equal zero.
So, statement 1 is NOT SUFFICIENT

Statement 2: x^2 = y^2
If we subtract y^2 from both sides, we get x^2 - y^2 = 0
If we factor the left side, we get (x+y)(x-y) = 0
From this, we can conclude that either x+y = 0 (in which case the answer to the target question is YES)
Or x-y = 0 (in which case the answer to the target question may not be YES)
So, statement 2 is NOT SUFFICIENT

Statements 1 + 2:
From statement 2 we get two possibilities:
case a) x+y=0 or case b) x-y = 0

However, statement 1 tells us that xy < 0, which means that x and y have different signs.
This helps us rule out case b) (x-y=0) since it's impossible for x and y to have different signs AND for x-y=0
This means that case a (x+y=0) must be true.
So, statements 1 and 2 combined are SUFFICIENT

Answer = C

Cheers,
Brent