is it E?

[mehaksal]

is x>y>

1. sqrt(x)> sqrt(y)
2. x^2 > y^2

[niketdoshi123]

is x>y>

1. sqrt(x)> sqrt(y)
2. x^2 > y^2

Statement 1:

let sqrt(x) = 3 and sqrt(y) = 2
So, x = 9 & y = 4 => x>y

let sqrt(x) = 1/2 and sqrt(y) = 1/3
So, x = 1/4 & y = 1/9 => x>y

sqrt of any number cannot be -ve. So we don't have to check for negative numbers.

Hence sufficient

Statement 2:

let x = 3 => x^2 = 9
and let y= 2 => y^2 = 4
here x^2>y^2 and x>y

Let x = -3 => x^2 = 9
and let y = 2 => y^2 = 4
here x^2>y^2 , but x<y

Hence insufficient

The correct answer is A

[mehaksal]

"sqrt of any number cannot be -ve. So we don't have to check for negative numbers. " ??

my confusion in this question is regarding the sqrt in 1)..why can't we consider negative sqrt here?

[niketdoshi123]

"sqrt of any number cannot be -ve. So we don't have to check for negative numbers. " ??

my confusion in this question is regarding the sqrt in 1)..why can't we consider negative sqrt here?

Ok..
Consider x =-9
what will be the square root of x?? It will be -3 right.. The answer is NO.. the sqrt(-9) will be 3i
where i = sqrt(-1).
3i is considered an imaginary number and imaginary numbers cannot be represented in a 1-dimensional number line... This concept is not tested on the GMAT. So you don't have to worry about it.

Also every non- negative real number has non- negative sqrt, called the principal sqrt.