Question

Six years before, the age of a mother was equal to the square of her son’s age. Three year hence, her age will be thrice the age of her son then. Find the present ages of the mother and son.

Answer 1

Hint:
We can assume the age of the mother to be x and age of the son to be y. Then we can consider the ages six years ago. Then we can form an equation with the given details in the question. Then we can consider the ages at 3 years later and form an equation with the given relation, then we can eliminate one variable by substitution. Then we can solve the quadratic equation to get one variable. Then we can re-substitute this value to get the other variable. The values of the variables will give the required ages.

Complete step by step solution:
Let x be the present age of the mother and y be the present age of the son.
It is given that six years before, the age of the mother was equal to the square of the son’s age. So, we can write this relation as an equation.
 \[ \Rightarrow x - 6 = {\left( {y - 6} \right)^2}\] … (1)
Now consider the case 3 years later.
 \[ \Rightarrow x + 3 = 3\left( {y + 3} \right)\]
On expanding the bracket, we get,
 \[ \Rightarrow x + 3 = 3y + 9\]
On simplification, we get,
 \[ \Rightarrow x = 3y + 6\] …. (2)
On substituting (2) in (1), we get,
 \[ \Rightarrow 3y + 6 - 6 = {\left( {y - 6} \right)^2}\]
On simplification, we get,
 \[ \Rightarrow 3y = {\left( {y - 6} \right)^2}\]
On expanding the RHS using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ , we get,
 \[ \Rightarrow 3y = {y^2} - 12y + 36\]
On rearranging, we get,
 \[ \Rightarrow {y^2} - 15y + 36 = 0\]
Now we can solve the quadratic equation by splitting the middle term.
 \[ \Rightarrow {y^2} - 12y - 3y + 36 = 0\]
On taking the common factors, we get,
 \[ \Rightarrow y\left( {y - 12} \right) - 3\left( {y - 12} \right) = 0\]
Again, on taking the common factors, we get,
 \[ \Rightarrow \left( {y - 3} \right)\left( {y - 12} \right) = 0\]
 $ \Rightarrow y = 3,12$
When $y = 3$ , the age six years ago will become $3 - 6 = - 3$ , which is negative. As age cannot be negative, we can reject this value of the variable.
 $ \Rightarrow y = 12$
On substituting in equation (2), we get,
 \[ \Rightarrow x = 3 \times 12 + 6\]
 \[ \Rightarrow x = 36 + 6\]
 \[ \Rightarrow x = 42\]

So, the present age of the mother is 42 and that of her son is 12.

Note:
While taking the variable, we must specifically declare them which variable represents the age of the mother and which gives the age of the son. After solving the quadratic equation, we will obtain 2 values for y and only one value is taken. One solution is rejected because the age before 6 years will become negative and age cannot be a negative value. We can also verify our answer by substituting the solution we got in the equations.