Question

One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son’s age. Find their present ages.

Answer 1

Hint: We will be using the concept of word problems to create equations with the help of conditions given in the question then we will use equations in two variables to further simplify the process and find the answer.

Complete step-by-step solution -
Now, we have been given that one year ago, a man was 8 times as old as his son.
So, we let the present age of man = x years.
The present age of son = y years.
Now, according to the question we have,
$\left( x-1 \right)=8\left( y-1 \right)..........\left( 1 \right)$
Also, it has been given that now the age of man is the square of his son. Therefore, according to the question we have,
\[x={{y}^{2}}.........\left( 2 \right)\]
Now, we will solve (1) and (2) to find the value of x and y from (1) we have the value of x in terms of y as,
$\begin{align}
  & x=1+8y-8 \\
 & =8y-7 \\
\end{align}$
So, we will substitute this in (2). So, we have,
$\begin{align}
  & 8y-7={{y}^{2}} \\
 & {{y}^{2}}-8y+7=0 \\
\end{align}$
Now, we will use the method of factorization to solve the quadratic equation. So, we have,
$\begin{align}
  & {{y}^{2}}-7y-y+7=0 \\
 & y\left( y-7 \right)-1\left( y-7 \right)=0 \\
 & \left( y-1 \right)\left( y-7 \right)=0 \\
\end{align}$
Either y – 1 = 0 or y – 7 = 0.
So, we have either y = 1 or y = 7. Now, for y = 1 we have $x={{\left( 1 \right)}^{2}}=1$, which is practically not possible as the age of both man and his son can’t be the same.
Therefore, we have the age of a son = 7 years.
The age of man $={{\left( 7 \right)}^{2}}=49years$.

Note: To solve these types of questions one must first convert the conditions given in the questions as an equation and then solve it for answer. Also it is important to note how we have eliminated the possibility of the age of the age of son = 1 year.