Is xy + xy < xy ? (1) x^2/y < 0 (2) x^3y^3 < (xy)^2

Hello Mun23,

Let me try answering this question for you. This question asks us whether 2xy<xy. Now the question we need to ask ourselves is when would this turn out to be true?

If x and y are both positive, then 2xy>xy.
If x or y =0, then 2xy=xy.
If either x or y is negative and the other is positive, then 2xy<xy. In short xy needs to be <0 or negative.

So, we need to find whether exactly one of those variables is positive and the other negative.

Now, 1 gives us x^2/y<0. Since, x^2 is always positive, this tells us that y is negative. However, we have no idea about the actual value of x.
x^2=some positive number
x could be a positive number or negative number. However, it cannot be 0. Hence, INSUFFICIENT.
Statement 2 gives us x^3*y^3<(xy)^2
Since (xy)^2 is always positive, we can divide the inequality by (xy)^2 without changing the sign.

Hence, (xy)<1. However, we still don't know their exact value. If xy<0, then one of the variables is positive and the other negative and this would be sufficient. However, if x or y=0, even then x*y<1. This does not give us conclusive proof that one of the variables is positive and the other negative(xy is negative). Hence, INSUFFICIENT.

From statement 1 we know that y<0. However, x can be negative or positive. However, x cannot be 0 as then x^2/y=0.
From statement 2 we know that x*y<1. We also know from 1) that y is negative. However, x could still be negative or positive.

Now, what if x=1/3 and y=-1, x^2/y=-1/9<0 and xy=-1/3<1. Is xy negative? Yes
What if x=-1/3 and y=-1. x^2/y=-1/9. However, xy=1/3<1. iS xy negative? No

Hence, INSUFFICIENT and the answer is E.

Hope this helps! Let me know if I can help you any further.