Question

The sum of the ages of a man and his son is $45$ years. Five years ago, the product of their age was four times the man's age at that time. Find their present ages.

Answer 1

Step 1: Determining the quadratic equation in $y$

Let the present age of the son be $y$ years.

Then the present age of the father is $(45-y)$ years.

As per the given condition, the ages of the man and the son are as follows:

	Present age in years	Age in years, five years ago
Man	$(45-y)$	$45-y-5=(40-y)$
Son	$y$	$(y-5)$

Step 2: Determining the ages of the father and the son

Given that five years ago, the product of their age was four times the man's age at that time.

$$\begin{gathered} (40-y)(y-5)=4(40-y) \\ \Rightarrow y-5=4 \\ \therefore y=9 \end{gathered}$$

So, $(45-y)=45-9=36$

Therefore, the present age of the son is $9$ years and his father is $36$ years.