Data Sufficiency

Is x > y?

(1) x^(1/2)> y

(2) x^3 > y

Stmt1:
x^(1/2) > y
Square both sides:
x > y^2 implies x > y. Sufficient.

Stmt2:
x^3 > y.
suppose x = 2, y = 4
x^3 = 8 is > y but x < y.
Suppose x = 2, y = 1
x^3 = 8 is > y and x > y. Insufficient.

Hence answer should be A.

I think the ans should be C....

In GMAT the square root of a number always refers to the positive root...So in this case we have to assume x to be a positive number

statement 1:
sqrt(x)>y
Let x=4 and y=1
then sqrt(4)>1
Let x=1 and y=4
then sqrt(1)<4
statement 1 is not sufficient

statement 2:
x^3>y
let x=2 and y=1 ......x^3>y
let x=1 and y=2 ......x^3<y
statement 2 is not sufficient

Combining both statements 1 and 2
x=2 y=1 we get x>y
x=1 y=2 we get x<y
x=1/4 y=1 we get x<y
x=16 y=-2 we get >y

As a general rule if both square root and cube of a number is greater than the other number,the number is always greater than the other number.

raju232007 wrote:I think the ans should be C....

In GMAT the square root of a number always refers to the positive root...So in this case we have to assume x to be a positive number

statement 1:
sqrt(x)>y
Let x=4 and y=1
then sqrt(4)>1
Let x=1 and y=4
then sqrt(1)<4
statement 1 is not sufficient

You cant take x=1 & y=4

because these values do not hold true for statement I

You dont have to prove the statement I, it is given as a fact.

statement 2:
x^3>y
let x=2 and y=1 ......x^3>y
let x=1 and y=2 ......x^3<y
statement 2 is not sufficient

You cant take x=1 & y=2

because these values do not hold true for statement 2

You dont have to prove the statement 2, it is given as a fact.

As a general rule if both square root and cube of a number is greater than the other number,the number is always greater than the other number.

for me answer is C
combining 1 and 2 we get that x is an integer, sqrt(x) and x^3 are greater than y, so x>y


OA?

OA is C

I agree with C now.
I didn't consider sqrt(x) could be a fraction and then what I posted earlier regarding Stmt1 would not hold true.