Is |xy|> (xy)^2

[uzair_baig]

Is |xy|> (xy)^2 ?

1. 0<x^2<1/4
2. 0<y^2<1/9

The OA is C. I don't quite understand why its C and not D.

[RonPurewal]

Is |xy|> (xy)^2 ?

1. 0<x^2<1/4
2. 0<y^2<1/9

The OA is C. I don't quite understand why its C and not D.

the left-hand side is the square root of the right-hand side. i.e., |xy| is the square root of (xy)^2.

so, a YES to this question means that, when |xy| is squared, it gets smaller.
if |xy| is between 0 and 1, the answer to the question will be YES.

the individual statements aren't sufficient because they don't impose any constraints at all on the other number.
for instance, statement (1) limits x to between 0 and 1/2 (or between -1/2 and 0) -- but, under that statement, y can be literally any number in the world.
so, if x is, say, 1/4 and y is another fraction, you're going to get a YES to the question.
but, if x is 1/4 and y is 1,000,000,000, then you're going to get a NO to the question.

same logic applies to the other statement, with the roles of x and y reversed.

once you get the two statements together, though, you know for sure that both x and y are fractions. this way |xy| is definitely less than 1, and so it will definitely get smaller when you square it.

[prashant.ranjan]

Greetings Ron,

I just wanted to clarify one doubt here. The question hasn't mentioned that x and y are different. Even combining (1) and (2), if we consider
x^2 = 1/25 (1/25<1/4). So x may be (+/- 1/5). Similarly if we consider y^2 = 1/25 (1/25< 1/9). So (y may be +/- 1/5).
Irrespective of the signs if we consider (since both sides are going to be positive anyways), we would get

|x.y| !> (x.y)^2

I know if this is a GMAT Prep question, then it's preposterous to question the correct answer. But if I would have got this question, I would have marked it (E) : (
Am I missing something here : (

Thanks
Prashant
An ardent fan of RON :) ...

[RonPurewal]

Greetings Ron,

I just wanted to clarify one doubt here. The question hasn't mentioned that x and y are different. Even combining (1) and (2), if we consider
x^2 = 1/25 (1/25<1/4). So x may be (+/- 1/5). Similarly if we consider y^2 = 1/25 (1/25< 1/9). So (y may be +/- 1/5).
Irrespective of the signs if we consider (since both sides are going to be positive anyways), we would get

|x.y| !> (x.y)^2

try again.
with your choice of numbers, |xy| is 1/25, but (xy)^2 is (1/25)^2, which is a lot smaller.

[prashant.ranjan]

Gosh!!!!!!
What i was thinking :P
Srry to take up your time..

Thanks and Regards
Prashant

[tim]

:)