If 1 were true, then we'd have to determine if 2x is greater than x^2. While this is true for numbers greater than 2 (just pick any number and you will see), it is not for numbers between 0 and 2. So 1 is insufficient.

2. We all know that a perfect square is greater than or equal to zero, which means that (x - y)^2 > = 0. This is equivalent to x^2 + y^2 - 2xy >= 0 . This means that x^2 + y^2 >= 2xy. Now, since x + y > x^2 + y^2, then, using the previous equation, we get that x + y > 2xy. Since both x and y are non-negative, then 2xy will be greater than xy, which ultimately leads us to x + y > xy.

So I'd go with the same answer.

DanaJ wrote:If 1 were true, then we'd have to determine if 2x is greater than x^2. While this is true for numbers greater than 2 (just pick any number and you will see), it is not for numbers between 0 and 2. So 1 is insufficient.

2. We all know that a perfect square is greater than or equal to zero, which means that (x - y)^2 > = 0. This is equivalent to x^2 + y^2 - 2xy >= 0 . This means that x^2 + y^2 >= 2xy. Now, since x + y > x^2 + y^2, then, using the previous equation, we get that x + y > 2xy. Since both x and y are non-negative, then 2xy will be greater than xy, which ultimately leads us to x + y > xy.

So I'd go with the same answer.

What if x = y = 0?

IMO: (1) does not work by itself since x = y = 0 and x = y = 1 give different answers. (2) means that either x or y are both fractions between 0 and 1, since that is the only way squaring can reduce the result. Adding a fraction (x + y) increases the result, but multiplying by a fraction (x y) decreases it. Thus x + y > x y under (2). Checking for zero gives a compatible result.

Well, since you have x + y > x^2 + y^2, then x = y = 0 is not possible, since both sides of the inequality are equal to zero. This is how I eliminated this case for stmt 2.

DanaJ wrote:Well, since you have x + y > x^2 + y^2, then x = y = 0 is not possible, since both sides of the inequality are equal to zero. This is how I eliminated this case for stmt 2.

Now fine

But still the statement asks whether 2x> X^2, substituting number from 0 to 2 & numbers greater than 2 would tell us that 2X could be equal or lesser than X^2 but never be greater, right?

So A not sufficient? Please advise.

GID09 wrote:But still the statement asks whether 2x> X^2, substituting number from 0 to 2 & numbers greater than 2 would tell us that 2X could be equal or lesser than X^2 but never be greater, right?

So A not sufficient? Please advise.

Very good point made in hurry. Check it out:

x 2 x x^2

0 0 0
1/2 1 1/4
1 2 1
3/2 3 9/4
2 4 4
5/2 5 25/4
3 6 9

and so on...

Did you notice something?

Thanks Sanju 09!

DanaJ wrote:If 1 were true, then we'd have to determine if 2x is greater than x^2. While this is true for numbers greater than 2 (just pick any number and you will see), it is not for numbers between 0 and 2. So 1 is insufficient.

2. We all know that a perfect square is greater than or equal to zero, which means that (x - y)^2 > = 0. This is equivalent to x^2 + y^2 - 2xy >= 0 . This means that x^2 + y^2 >= 2xy. Now, since x + y > x^2 + y^2, then, using the previous equation, we get that x + y > 2xy. Since both x and y are non-negative, then 2xy will be greater than xy, which ultimately leads us to x + y > xy.

So I'd go with the same answer.

nice approach.

GID09 wrote:Thanks Sanju 09!

Pleasure is mine GID09. Please read the above table contents pretty carefully, formatting is still a problem here, please read column 1 for x, 2 for 2 x and 3 for x^2, and let me please know what you really noticed out of it.

Thanks