Question: R, S and T are 3 consecutive odd integers.
Column A

Column B

R + S + 1

S + T – 1

Solution: Let R = 1, then S = 3, T = 5
Column A= R + S + 1 = 1 + 3 + 1 = 5.
Column B = S + T – 1 = 3 + 5 – 1 = 7.
Column B is Greater.
But let R = 1, then S = 3, T =  5
Column A = R + S + 1 = 1 – 3 + 1 = 3.
Column B = S + T – 1 = 3 – 5 – 1 = 9.
Now, column A is greater.
So, answer is option (D).