If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

i know this problem has been solved in this forum few times but none of them were able 2 solve my confusion.
For the second part r>s

If i plug in negative number r=-2, s=-3
then -5>2*(any value of t for which equation is true, for eg -4)
-5>-8
but when we compare r and t we find -5<-4. (You have considered r = -2 and not -5, so this would be wrong)

Then how can the solution be D? i guess it should be A.
Please correct me.

this one doesn't require values to be plugged in.
we can just change the rules and solve the inequation

give r + s > 2t
1) t > s
=> -s > -t
add this with given
r+s-s>2t -t
r > t
hence sufficient.
2) r > s
=> -s > -r
add this with given
r + s -s > 2t - r
2r > 2t
r > t
hence sufficient.

even if you keep values and solve, the inequalities wont change and you will get the same answer D.

user123321

deleted the post.

user123231

I think 'nidhis.1408' used

r = -2
s = -3 and
t = -4

You have, r+s = -2-3 = -5

and 2t = -8.

But instead of comparing 'r' and 't', (i.e -2 and -4), nidhis has compared (r+s and t) i.e( -5 and -4). Hence the misleading answer.

@shankar.ashwin you are absolutely right. i made d mistake of comparing r+s to t thats why i couldn't conclude that both the statement are sufficient.
Thanks a lot

r+s > 2t
r-t > t-s ---- 1

a. t> s = t-s > 0
thus in -- 1

r-t > t-s > 0 meaning r-t > 0 = r > t sufficient.

b r> s = r-s > 0 ----2

adding --1 and --2

2r > 2t meaning r> t hence sufficient.

thus D it is.