M7MBA wrote:Is \(x+y>0 ?\)
(1) \(xy>0\)
(2) \(x^3y^2>0\)
Statement 1:
Here, x and y must have the SAME SIGN.
If x and y are both positive, then x+y > 0.
If x and y are both negative, then x+y < 0.
INSUFFICIENT.
Statement 2:
The inequality implies that x and y are both NONZERO.
Since the square of a nonzero value must be positive, we can safely divide both sides by x²y²:
x³y²/x²y² > 0/x²y²
x > 0
No information about y.
If x=1 and y=1, then x+y > 0.
If x=1 and y=-2, then x+y < 0.
INSUFFICIENT.
Statements combined:
Since x>0 and x and y have the same sign, y>0.
Thus, x+y > 0.
SUFFICIENT.
The correct answer is
C.
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