yalanand wrote:If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

We know that r+s>2t, so r-t>t-s

from (1): t>s => t-s>0 => r-t>t-s>0 => r>t so (1) alone is sufficient.

from (2): r>s => r+s>2s.............(a) and we have r+s>2t............(b)
(a)+(b) => 2r+2s>2s+2t => r>t, so (2) alone is also sufficient.

My answer is (D).

agree with billzhao... if you plug numbers in, you can find the same answer: if you satisfy the equation and the rules it gives you, no matter what r>t.

yalanand wrote:If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

Given r + s > 2t.

Statement 1: t > s i.e s < t

Since r + s > 2t and s < t it is obvious that r > t.

Hence statement 1 alone is sufficient.

Statement 2: r >s

Given r + s > 2t

Now if r < = t, then obviously s > t.

But in that case r < s thus contradicting statement 2.

So r > t.

Hence statement 2 alone is sufficient.

So D must be the answer.