Question

The ages of Kavya and Karthik are 11 years and 14 years respectively. In how many years will the product of their ages be 304?

Answer 1

Hint: Suppose the product of ages will be 304 after ‘n’ years. Now form a quadratic with the help of the given equation. Hence, using factoring or quadratic formula, find the value of ‘n’.

Complete step-by-step answer:

Here, ages of Kavya and Karthik are given as 11 years and 14 years. Now, we need to determine after how many years, the product of their ages will be 304.
Let us suppose the product of the ages of Kavya and Karthik is 304 after ‘n’ years from their present age.
Therefore, the ages of Kavya and Karthik will be (11 +n) and (14 + n) years, respectively after ‘n’ years from now.
Hence, we can form an equation by using the product of ages given in the question by using the product of ages given in the question and the supposed ages as
(11 + n) (14 + n) = 304 ……….. (i)
Let us multiply the above equation, we get
$\begin{align}
  & 11\times 14+11n+14n+{{n}^{2}}=304 \\
 & \Rightarrow {{n}^{2}}+25n+154=304 \\
 & \Rightarrow {{n}^{2}}+25n-150=0........(iii) \\
\end{align}$
Now, we can factorize the above equation and get the value of ‘n’.
So, we can split ’25’ as ‘30’ and ‘-5’ to get the same multiplication ‘-150’of constant term and coefficient of ${{x}^{2}}$.
Hence, equation (ii), can be written as
${{n}^{2}}+30n-5n-150=0$
Now take ‘n’ as common from the first two terms and ‘-5’ from the last two terms. Hence, we get
n(n + 30) -5(n +30) = 0
(n - 5) (n + 30) = 0
Now equate n – 5 and n + 30 to zero to get the values of n.
Hence, we get
n – 5 = 0or n + 30 = 0
n = 5 or n = -30
An n = -30 is not possible because we are talking about the future and if we put n = -30 with ‘11+n’ and ‘14+n’, so we will get imaginary ages of Kavya and Karthik.
Hence, n = 5 is the correct answer.
So, after 5 years from present the product of the ages of Kavya and Karthik would be 304.

Note: One can solve the quadratic ${{x}^{2}}+25n-50=0$ by using the quadratic formula for getting roots as well which is given as
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$ for $A{{x}^{2}}+Bx+C=0$.
One can get value of ‘n’ by comparing the equation
\[\left( n+11 \right)\left( n+14 \right)=16\times 19\]
Here the difference between terms n + 11 and n + 14 is 3 and between 16 and 9 as well.