Question

The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than 3 times the age of her son. Find their present age?

Answer 1

Hint:First assume the present age of mother is x years and son be y years then apply the two conditions given in the question from where you will get two equations. Now, you have two equations and two unknowns solve them and you will get the value of x and y.

Complete step-by-step answer:
Let us assume the present age of mother is x years. And the present age of the son is y years.
It is given that the age of the mother is twice the square of the son’s age. Writing this statement in the form of equation we get,
$x = 2y^2………. Eq. (1)$
After 8 years, the age of the mother will be (x + 8) years and the age of son will be (y + 8) years. Then the relation between their ages is given as the age of the mother is 4 years more than three times the age of son.
$(x + 8) = 4 + 3(y + 8)$
$\Rightarrow x + 8 = 4 + 3y + 24$
$\Rightarrow x = 3y + 20$
Substituting the above value of x in eq. (1) we get,
$3y + 20 = 2y^2$
$\Rightarrow 2y^2 – 3y – 20 = 0$
The above equation is quadratic in y so solving the above equation we get,
$\begin{align}
  & 2{{y}^{2}}-8y+5y-20=0 \\
 & \Rightarrow 2y\left( y-4 \right)+5\left( y-4 \right)=0 \\
 & \Rightarrow \left( 2y+5 \right)\left( y-4 \right)=0 \\
\end{align}$
The solutions of y are $4$ and$-\dfrac{5}{2}$. And as age cannot be negative so we are rejecting the negative solution of y and only considering the value of y as 4.
 So, the present age of a son is equal to 4 years.
Substituting the value of y = 4 in the eq. (1) we get,
$x = 3(4) + 20$
$\Rightarrow x = 32$
So, the present age of a mother is $32$ years.
Hence, the present age of son is equal to 4 years and the present age of mother is 32 years.

Note: Always verify the solutions that you are getting from the quadratic equation by satisfying those solutions in the conditions given in the question. Like in the question we get two values of y one is 4 and other is $-\dfrac{5}{2}$and as we know that age of a person cannot be negative and in a fraction so we have rejected the solution$-\dfrac{5}{2}$and y = 4 is the only solution that we have considered.