question stem: r + s > 2t
stmt(1) t > s
add both the inequalities
r + s + t > 2t + s
=> r + s - s > 2t - t (just to explain the middle step)
=> r > t
(1) is suffiecient
now, question stem: r + s > 2t
r> s
adding both the inequalities
2 r + s > 2t + s
=> 2r + s - s > 2t
=> 2r > 2t
r>t (since 2 is a positive number, no change of sign)
(2 is sufficient)
ans : D each stmt is sufficient