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
