i get A

i mean E

Could you post ur solution?

This is what I think the answer is

gmat_enthus wrote:Guess I missed to post this one;

Q. Is xy < x^2*y^2?

1) xy>0
2) x+y=1

First simplify to xy < (xy)^2 using law of exponents

With statement (1), you know that either both x and y are negative or both x and y are positive. Otherwise their product could not be positive.
However, even within this space the answer to the ineuqality is ambiguous since for -1 < x < 1 and -1 < y < 1, the inequality does not hold, but for two negative numbers or two positive numbers greater than 1 or less than -1, it does hold.

With statement (2) you know that either x and y are both greater than zero and less than one such that their sum equals 1 (e.g. - 1/3 and 2/3), or you know that they are two numbers (one positive and one negative) where the positive number has an absolute value 1 greater than the negative number. This statement too is ambiguous since as in the example given the product of 1/3 and 2/3 is greater than their product squared but the product of 8 and -7 is less than their product squared (-56 < (-56)^2).

Take the two together and you see that a non-negative product of two numbers whose sum equals 1 only allows for 0 < x < 1 and 0 < y < 1, and in this case, the inequality is always false.

Answer: E

** frustrating bit - just getting started with DS type Q questions and theya re kicking my butt. Took me 20 minutes to reason through it and write these words down

gmatleyFool wrote: Take the two together and you see that a non-negative product of two numbers whose sum equals 1 only allows for 0 < x < 1 and 0 < y < 1, and in this case, the inequality is always false.

Answer: E

** frustrating bit - just getting started with DS type Q questions and theya re kicking my butt. Took me 20 minutes to reason through it and write these words down

First of all welcome to the forums gmatleyFool!

great explanation there.
you must know that this is one of the toughest problems you d encounter in GMAT test.

i want you to have a look at it once again. 8)
just check it again

Hi aim-wsc

Thanks for the welcome.. looking forward to more participation in the future.

I checked my answer.. I meant to say C
When taken together they are enough, each alone is not.

There.. is that better?

thats better.

DS problems are just like that.
be careful when solving those.

just another solution for this one.
xy < x^2y^2 => xy(1-xy) < 0 let xy=A so we have A(1-A)<0
and either A<0 and A<1 or A>0 and A>1 but from the statements we have A=xy>0 but since x+y=1 then A=xy cannot be > 1.
Therefore inequality never holds