Ages 68Q

68. The ages of two persons differ by 16 years. If 6 years ago, the elder one be 3 times as old as the younger one, find their present ages.

A.  14 years      B.  17 years

 C.  24 years     D.  29 years

Answer:

Let the age of the younger person be x years.

Then, age of the elder person = (x + 16) years.

Therefore         ‹=› 3 (x - 6) =(x + 16 - 6)

‹=› 3x-18 = x+10

‹=› 2x = 28

x= 14 years.