Ten years ago, a man was seven times as old as his son. Two years hence, twice his age will be equal to five times the age of his son. What is the present age of the son ?

112 years
213 years
314 years
415 years

Answer:- 1

Explanation:-

Solution:

Let son's age 10 years ago be x years.
Then, man's age 10 years ago = 7 years
Son's present age = (x + 10) years,
Man's present age = (7x + 10) years

\begin{aligned} ∴ 2 [(7 x + 10) + 2] = 5 [(x + 10) + 2] \\ \Rightarrow 2 (7 x + 12) = 5 (x + 12) \\ \Rightarrow 14 x + 24 = 5 x + 60 \\ \Rightarrow 9 x = 36 \\ \Rightarrow x = 4 \end{aligned}

$\eqalign{ & \therefore {\text{2}}\left[ {\left( {7x + 10} \right) + 2} \right]{\text{ = 5}}\left[ {\left( {x + 10} \right) + 2} \right] \cr & \Rightarrow 2\left( {7x + 12} \right) = 5\left( {x + 12} \right) \cr & \Rightarrow 14x + 24 = 5x + 60 \cr & \Rightarrow 9x = 36 \cr & \Rightarrow x = 4 \cr}$
Son's present age = (x + 10) years
= (4 + 10) years
= 14 years