Chapter 6.1, Problem 16P

Five years from now, a father’s age will be three times his son’s age, and 5 years ago, he was seven times as old as his son was. What are their present ages?

To determine

To find: The present age of father and son.

Answer to Problem 16P

The present age of son is 10 years and present age of father is 40 years.

Explanation of Solution

Given information: Five years from now, a father’s age will be three times his son’s age, and 5 years ago, he was seven times as old as his son was.

Calculation: Let the present age of son be x years.

Then, five years from now, age of son will be $(x+5)$ years and age of father will be $3(x+5)$ years. So, present age of father is $3(x+5)-5$ years.

Again, five years ago, age of son was $(x-5)$ years and age of father was $7(x-5)$ years. So, present age of father is $7(x-5)+5$ years.

According to the problem,

  $\begin{array}{l}3(x+5)-5=7(x-5)+5\\ \Rightarrow 3x+15-5=7x-35+5\\ \Rightarrow 3x+10=7x-30\\ \Rightarrow 7x-3x=10+30\\ \Rightarrow 4x=40\\ \Rightarrow x=10\end{array}$

  $\therefore$ The present age of son $=10$ years

  $\therefore$ The present age of father

  $\begin{array}{l}=3(x+5)-5\\ =3(10+5)-5\\ =3\times 15-5\\ =45-5\end{array}$

  $=40$ years

Thus, the present age of son is 10 years and present age of father is 40 years.