Five years ago a man's age was 3 more than 4 times his son's age. After 3 years from now, the man will be 6 less than thrice his son's age. After how many years will the sum of their ages be 80?

1. Let man = M, son = S now. (M−5) = 4(S−5) + 3 ⇒ M = 4S − 12. 2. (M+3) = 3(S+3) − 6 ⇒ M = 3S; so 3S = 4S − 12 ⇒ S = 12, M = 36. 3. Sum now = 48; need 48 + 2y = 80 ⇒ y = **16** years. _Source: IBPS RRB Officer Scale I Prelims PYQ (Examrace), Q.67_