Question

Divide 15 into two parts such that the sum of their reciprocals is $\dfrac{3}{10}$.

Answer 1

Hint: Assume the first part as x and hence calculate the second part by subtracting x from 15. Take the reciprocal of these parts by dividing 1 by them individually and consider their sum equal to the given fraction $\dfrac{3}{10}$. Form a quadratic equation in x and solve it using the middle term split method to get the answer.

Complete step by step answer:
Here we have been asked to divide 15 into two numbers such that when we take the sum of their reciprocals we get $\dfrac{3}{10}$.
Now, let us assume the first part as x, therefore the second part will be given by subtracting x from 15, so we get the second part as 15 – x. Further, we need to consider the sum of the reciprocals of these two parts. We know that the reciprocal of a number is 1 divided by that number, therefore the reciprocal of x is $\dfrac{1}{x}$ and that of 15 – x is $\dfrac{1}{15-x}$. Equating the sum with $\dfrac{3}{10}$ we get,
$\begin{align}
  & \Rightarrow \dfrac{1}{x}+\dfrac{1}{15-x}=\dfrac{3}{10} \\
 & \Rightarrow \dfrac{15-x+x}{x\left( 15-x \right)}=\dfrac{3}{10} \\
 & \Rightarrow \dfrac{15}{x\left( 15-x \right)}=\dfrac{3}{10} \\
\end{align}$
Cancelling the common factors and cross multiplying the terms we get,
$\begin{align}
  & \Rightarrow \dfrac{5}{15x-{{x}^{2}}}=\dfrac{1}{10} \\
 & \Rightarrow 15x-{{x}^{2}}=50 \\
 & \Rightarrow {{x}^{2}}-15x+50=0 \\
\end{align}$
Now, using the middle term split method to factorize the quadratic expression we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}-10x-5x+50=0 \\
 & \Rightarrow x\left( x-10 \right)-5\left( x-10 \right)=0 \\
 & \Rightarrow \left( x-5 \right)\left( x-10 \right)=0 \\
\end{align}$
Substituting each term equal to 0 we get,
$\Rightarrow \left( x-5 \right)=0$ or $\left( x-10 \right)=0$
\[~\Rightarrow \] x = 5 or x = 10
Therefore, the corresponding two values of 15 – x will be 10 or 5.
Hence, the two parts in which 15 must be divided is 5 and 10.

Note: You may think that why we are getting two values of x here. Actually there is only one solution i.e. 5 and 15 but we are getting two values because we have two variables namely x and 15 – x. So in case x is 5 then 15 – x is 10 and in case x is 10 then 15 – x is 5. Both the cases are true so we can consider both of them.