jaspreetsra wrote:If x and y are integers, is x>y?
1) x+y>0
2) y^x<0
1) x+y>0; clearly not sufficient
2) y^x<0; y may be any -ve integer and x may be any -ve or +ve odd integer. Not sufficient
What next???
Help please!
Statement 1:
It's possible that x=2 and y=1, in which case x>y.
It's possible that x=1 and y=1, in which case x=y.
INSUFFICIENT.
Statement 2:
y^x < 0 implies that y<0 and that x is ODD.
It's possible that y=-1 and x=1, in which case x>y.
It's possible that y=-1 and x=-1, in which case x=y.
INSUFFICIENT.
Combined:
Linking together x+y > 0 and 0 > y, we get:
x+y > 0 > y
x+y > y
x > 0.
Linking together x > 0 and 0 > y, we get:
x > 0 > y
x > y.
SUFFICIENT.
The correct answer is
C.
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