Yes, the answer is B.
Actually, what I wanted is to solve this fastest possible and most accurate possible. Plug in numbers could be quite tedious (especially if one do not know what are the possible ranges of numbers to plug in).
My working is this:
Now, before we proceed further, we must understand that from answer choices (A) to (E), there is none that permits us to choose "None of the above"! In addition, if condition (I): X>1 is true, then condition (II): X^2>1 must also be true.
So, if we look at the answer choices again, we can eliminate (C) and (D). And among (A), (B) and (E), we just have to prove that (A) is not NECESSARILY true to eliminate BOTH (A) and (E), leaving us with (B). So, let's attempt to prove that (A) or (III): XY>1 is not necessarily true (actually at this juncture, we have to keep our fingers crossed and hope that we CAN really prove that (A) is not necessarily true).
Try plug in numbers that satisfied XY<1: say X= 2 (notice that I use X=2 to satisfy condition (I)!!!! I shouldn't be using a number X such that X<1)and Y=1/3. So, plug in both number and you will notice that these 2 numbers satisfy both (1) and (2)! Effectively, we have shown that (III) is not necessary true, and therefore (A) and (E) are out too!
We have X^2*Y>1 and Y^2*X<1
=>X^2*Y>1 and -Y^2*X>-1 (the > becomes < when the sign changes).
In Inequalities you can add these two equations PROVIDED the > or < signs are same for both the equations.
So we now have X^2*Y - Y^2*X > 0
=>XY*(X-Y) > 0
Now, it is important, at this juncture to note that the question statement already tells us that X>0 and Y>0!!! So, this inequality only provides us with the additional info that (X-Y)>0 must be true.
X>0 and Y>0 => XY>0. However, this does not mean that you can immediately conclude that XY>1 is not NECESSARILY true to satisfy (1) and (2). You still have to prove that there are numbers X,Y that satisfy 0<XY<1, in order to prove that XY>1 is not a necessary condition.
Venksume: what do you think? do you agree with me?