This problem is solvable by a combination of concepts and plugging in.knight247 wrote:Is X>Y
(1) x/a < y/a
(2) x > 2y
(1) really depends on the value of a. If a is positive, then x is smaller than y: try a=1, x=2, y=3 (x is smaller than Y) so that 2/1 < 3/1.
But if a is negative (for example a=-1), then suddenly you need x=3, y=2 (x is greater than y) to make 3/-1 =-3 smaller than 2/-1 = -2.
Insufficient
(2) x can obviously be greater than y if x and y are positive. Can we find a counter example, where x is not greater than y? How could x be greater than 2y, but less than y?
Only if y is somehow greater than 2y: y>x>2y -->y>2y --> 0>y
So if x and y are negative (for example, x=-3, y=-2), then x is indeed greater than 2y, but smaller than y. We have examples either way, one where x is greater and one where x is smaller than y, so insufficient.
Combined:
Take x and y as positives: x=5, y=2 works for stat. (2). Does it work for (1) as well? sure, if a is negative.
Take x and y as negatives where x is smaller than y: x=-3, y=-2 works for (2). Does it work for (1)? Sure, if a is positive, then then -3/a will be smaller than -2/a.
Thus, we can still find examples either way; one where x is greater than y, one where x is smaller than y. Still insufficient. The answer is E.