sqrt(x)> yselango wrote:Is x > y?
(1) sqrt(x)> y
(2) x^3 > y
OA C
let x=16, y =3
4>3
16>3....YES
Let x=0.25, y=0.4
0.5>0.4
0.25<0.4...NO
INSUFF
x^3 > y
let x=2, y=1
8>1
2>1....YES
let x=-1/2,y=-1/4
-1/8>-1/4
-1/2<-1/4...NO
INSUFF
combining:
X can be positive or negative or fraction or 0
X cannot be negative because sqrt(x) will not hold.
If x is 0, then y has to be negative to satisfy both the eqtns. hence X>Y
Now, If X is fraction, then stmt1 will fail. So, we cant input fractions.
Now, we are only left with positives. For positives, answer in oth the statements is YES.
Hence pick C.
