Question 6: The sum of the ages of the father and son is \(72\) years. Six years ago, the father’s age was \(2\) times the age of the son. What was son’s age six years ago?
Solution:
Let
\[
\begin{align*}
Father’s\ age\ &=\ F\\
Son’s\ age\ &=\ S\\
\end{align*}
\]
Six years ago
\[
\begin{align*}
Father’s\ age\ &=\ F-6\\
Son’s\ age\ &=\ S-6\\
\end{align*}
\]
According to the given conditon of the question, we have
\[
\begin{align*}
S+F&=72 \_\_\_ (1)\\
F-6&=2( S-6) \_\_\_( 2)
\end{align*}
\]
Frome Equation \((1)\)
\[
\begin{align*}
S+F&=72\\
F&=72-S\\
\end{align*}
\]
Putting value of \(F\) in equation \((2)\), we have
\[
\begin{align*}
72-S-6&=2S-12\\
-S-2S&=-12-72+6\\
-3S&=-78\\
S&=\frac{-78}{-3}\\
S&=26\\
\end{align*}
\]
So, Son’s age was \(20\) years six years ago.