grandh01 wrote:Is x - y > r - s?
(1) x > r and y < s?
(2) y = 2, s = 3, r = 5, and x = 6.

OA is D
Can someone explain A

(1) x > r and y < s
x > r and s > y
Adding both the inequalities, we get (x + s) > (r + y)
or (x - y) > (r - s), which answers the question as "yes"; SUFFICIENT.

(2) y = 2, s = 3, r = 5, and x = 6
x - y = 6 - 2 = 4
r - s = 5 - 3 = 2
Here again, (x - y) > (r - s), which answers the question as "yes"; SUFFICIENT.

The correct answer is D.

grandh01 wrote:Is x - y > r - s?
(1) x > r and y < s?
(2) y = 2, s = 3, r = 5, and x = 6.

OA is D
Can someone explain A

Adding the inequalities works well here. Or you can notice that Statement 1 guarantees that x - r is positive, and that y - s is negative. Positive numbers are always greater than negative ones, so x - r must be greater than y - s, and we have

x - r > y - s
x - y > r - s

which is what we wanted. So Statement 1 is sufficient.

Statement 2 is clearly sufficient without doing any work, since if you know the precise values of all of your unknowns, of course you can answer absolutely any question at all about those unknowns. So the answer is D.