Ten years ago, the age of father was twelve times as old as his son and after ten years, the age of father will be twice as old as his son will be then. Now find the present ages.

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Ten years ago, the age of father was twelve times as old as his son and after ten years, the age of father will be twice as old as his son will be then. Now find the present ages.

Let’s the present ages of the father as a years and that of his son’s age as b years.

According to the question,,

After 10 years, the age of father will be $(a+10)$ years and son’s age will be $(b+10)$ years.

So, the equation is

$a+10=2(b+10)$

$a-10=2b+20$

$a-2b-10=0$ ……… (i)

Also again from the question it’s given as,

Ago 10 years, the age of father was $(a-10)$ years and the age of son was $(b-10)$ years.

therefore, the relation between their 10 years ago is given below

$a-10=12(b-10)$

$a-10=12b-120$

$a-12b+110=0$ ……… (ii)

Therefore, by solving (i) and (ii), we get the solution

Using cross-multiplication method , then

$\begin{array}{l} \frac{a}{{( - 2) \times 110 - ( - 12) \times ( - 10)}} = \frac{b}{{1 \times 110 - 1 \times ( - 10)}} = \frac{1}{{1 \times ( - 12) - 1 \times ( - 12)}}\\ \frac{a}{{ - 220 - 120}} = \frac{{ - b}}{{110 + 10}} = \frac{1}{{ - 12 + 2}}\\ \frac{a}{{ - 340}} = \frac{{ - b}}{{120}} = \frac{1}{{ - 10}}\\ \frac{a}{{340}} = \frac{b}{{120}} = \frac{1}{{10}} \end{array}$