Is x>y?
1.sqrt(x)>y
2.x^3>y
The ans is C..Can you explain this one?
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
A tricky one
This topic has expert replies
-
raju232007
- Master | Next Rank: 500 Posts
- Posts: 133
- Joined: Sat Dec 29, 2007 2:43 am
- Thanked: 12 times
A) X could be a +ve fraction, in which case sqrt(X) maybe > Y.
x=16 y=1/5 sqrt(X) > Y, but X < Y.
X and Y could be +ve integers, in which case X > Y. Not Sufficient.
We don't have to bother about -ve numbers here as sqrt is involved.
B) X and Y could be +ve integers, in which case X > Y.
X can be a -ve fraction. X =-1/2 Y=-1/5 X^3>Y, but X < Y.
If Y is a -ve integer, X could be a -ve fraction for this condition and X > Y.
Not sufficient.
A and B tell you that X and Y have to +ve integers and X > Y.
x=16 y=1/5 sqrt(X) > Y, but X < Y.
X and Y could be +ve integers, in which case X > Y. Not Sufficient.
We don't have to bother about -ve numbers here as sqrt is involved.
B) X and Y could be +ve integers, in which case X > Y.
X can be a -ve fraction. X =-1/2 Y=-1/5 X^3>Y, but X < Y.
If Y is a -ve integer, X could be a -ve fraction for this condition and X > Y.
Not sufficient.
A and B tell you that X and Y have to +ve integers and X > Y.
