The age of a father 10 years ago was thrice the age of his son. 10 years hence, the father's age will be… · ABC26GN1815
Subject: General Aptitude · Chapter: Problems on Ages · Exam: P.C.S. 2009
The age of a father 10 years ago was thrice the age of his son. 10 years hence, the father's age will be twice that of his son. The ratio of their present ages is
(a)8 : 5
(b)7 : 3
(c)9 : 5
(d)5 : 2
Answer
No answer key was printed for this question.
Explanation
Let son's age 10 years ago be $x$ years. Then, father's age 10 years ago $=3 x$ years. Son's age now $=(x+10)$ years, Father's age now $$=(3 x+10) \text { years. }$$ $(3 x+10)+10=2[(x+10)+10]$ $$\begin{aligned} & \Rightarrow 3 x+20=2(x+20) \\ & \Rightarrow 3 x+20=2 x+40 \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.