There are four friends in a group: A, B, C, and
D. What is the present average age of the friends? Statement I: The present age of D is one-third the age of A, and the present age of B is 4 years less than the age of
A. The average age of D and C is 16 years, and D is 24 years younger than
B. Statement II: Seven years ago, D was one-fifth the age of
A. The difference between the age of B and the age of D is the same as the difference between the age of A and the age of
C. The average age of C and B is 28 years.

ATQ, Let, age of 'D' = 'd' years Present age of 'A' = 'f' years Present age of 'B' = 'm' years Present age of 'C' = 's' years Statement-I: So, 3d = f f – m = 4 f = 4 + m d + s = 32 m – d = 24 m = d + 24 Since 3d = 4 + m 3d = 4 + d + 24 2d = 28 d = 14 years f = 3d = 42 years m = 14 + 24 = 38 years s = 32 – 14 = 18 years Average age of friends = (14 + 18 + 42 + 38)/4 = 28 years So, data in statement-I alone is sufficient to answer the question. Statement-II: So, 5 × (d – 7) = (f – 7) 5d – 35 = f – 7 5d – 28 = f And, f – s = m – d f + d = m + s m + s = 28 × 2 = 56 m + s + f + d = 56 + 56 = 112 Average age of friends = 112/4 = 28 years So, data in statement-II alone is sufficient to answer the question.