Question

Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as he was then. Find their present ages.

Answer 1

Hint – Here we will proceed by assuming the present ages of father and son as variables and as per given conditions in problem form the linear equations in two variables. By using linear equations solving methods we will get our answer.

Complete step-by-step solution -
Let the present age of father $ = x$ years.
And the present age of his son $ = y$ years.
Therefore, According to first condition: -
$x - 10 = 12\left( {y - 10} \right)$
$\Rightarrow x - 10 = 12y - 120$
$\Rightarrow x - 12y = - 120 + 10$
$\Rightarrow x - 12y = - 110$ ….. (1)

Therefore, According to second condition: -
$x + 10 = 2\left( {y + 10} \right)$
$\Rightarrow x + 10 = 2y + 20$
$\Rightarrow x - 2y = 20 - 10$
$\Rightarrow x - 2y = 10$ ….. (2)

Therefore, subtracting equation (2) from (1)
$(x - 12y = - 110)$ - $(x - 2y = 10)$
On subtracting, we will get
$\Rightarrow {- 10y} = - 120$
$\Rightarrow y = \dfrac{{ - 120}}{{ - 10}}$
 $\Rightarrow y = 12$

Therefore, put $y = 12$ in equation (1).
$ x - 12y = - 110$
$\Rightarrow x - 12\left( {12} \right) = - 110$
$\Rightarrow x - 144 = - 110$
$\Rightarrow x = - 110 + 144$
$\Rightarrow x = 34$
Therefore, the present age of the father is x = 34 years and the present age of y = 12 years.

Note – Whenever we come up with this type of problem, one must know that there are three ways to solve systems of linear equations in two variables : graphing method, substitution method and elimination method. (Here substitution method).