The average age of A, B, C and D is 40 years. The present age of B is 90% of the present age of A and the present age of D is 40%

The average age of A, B, C and D is 40 years. The present age of B is 90% of the present age of A and the present age of D is 40% of the present age of C. If 10 years hence the age of B is 24 years less than the age of C at that time, then find the sum of the present ages of B and D?

A.            50 years

B.            40 years

C.            70 years

D.            60 years

E.            90 years

 Sol:

A + B + C + D = 40 * 4 = 160

B = 90/100 * A

B/A = 9/10

Let B = 9k, A = 10k

D = 40/100 * C

D/C = 2/5

C – B = 24

C = 24 + 9k

D = 2/5 * (24 + 9k)

9k + 10k + 24 + 9k + 48/5 + 18k/5 = 160

k = 4

Present age of B = 9 * 4 = 36

Present age of D = 2/5 * (24 + 36) = 24

Sum of the ages of B and D = 36 + 24 = 60 years