Data Sufficiency

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

(1) t > s

(2) r > s

OA after some discussion

chaitanya.mehrotra wrote:If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

OA after some discussion

IMO A

given:

r + s > 2t => r - t > t - s --- (1)

from A we get, t -s > 0

putting this result in eqn 1 we get r - t > 0 => r > t SUFFICIENT

Op B

r > s =>

r + s > 2r or r + s < 2r ---> putting these two results in the below given eqn u get two different results and hence INSUFFICIENT

Given: r + s > 2t

hi all, i want to believe that the answer is D
(1)r+s>2t, and t>s, if we sum both we get
r+s+t>2t+s. canceling s, left us with r+t>2t. and finally r>t suff
(2)the same way
r+s>2t, and r>s, again sum, 2r+s>2t+s, cancel s and left with 2r>2t, r>t also suff

IMO D
I plugged in these numbers. All satisfied the rule!

1.t>s -->0>-1, -2>-8, 1/2>1/4, 5>10
2.r>s ->0>-1, -2>-8, 1/2>1/4, 5>10

Anurag@Gurome wrote:

chaitanya.mehrotra wrote:If r + s > 2t, is r > t ?

(1) t > s
(2) r > s

Statement 1: t > s ---> (t + t) > (s + t) ---> 2t > (s + t)
Hence, (r + s) > 2t > (s + t) ---> r > t

Sufficient

Statement 2: r > s ---> (r + r) > (r + s) ---> 2r > (r + s)
Hence, 2r > (r + s) > 2t ---> r > t

Sufficient

The correct answer is D.

How come we reach r > t from (r + s) > 2t > (s + t) and same for 2r > (r + s) > 2t ---> r > t

Please help me in understanding this.