Father's age is three times the sum of age of his two children.

Solution:

Let the present age of father be x years and the present ages of his two children’s be y and z years.

The present age of father is three times the sum of the ages of the two children’s. Thus, we have

$x=3(y+z)$

$\Rightarrow y+z=\frac{x}{3}$

After 5 years, father's age will be $(x+5)$ years and the children's age will be $(y+5)$ and $(z+5)$ years. Thus using the given information, we have

$x+5=2\{(y+5)+(z+5)\}$

$\Rightarrow x+5=2(y+5+z+5)$

$\Rightarrow x=2(y+z)+20-5$

 

$\Rightarrow x=2(y+z)+15$

So, we have two equations

$y+z=\frac{x}{3}$

$x=2(y+z)+15$

Here x, y and z are unknowns. We have to find the value of x.

Substituting the value of $(y+z)$ from the first equation in the second equation, we have

By using cross-multiplication, we have

$x=\frac{2 x}{3}+15$

$\Rightarrow x-\frac{2 x}{3}=15$

$\Rightarrow x\left(1-\frac{2}{3}\right)=15$

$\Rightarrow \frac{x}{3}=15$

$\Rightarrow x=15 \times 3$

 

$\Rightarrow x=45$

Hence, the present age of father is 45 years.