Question

The difference of two natural numbers is 5 and the difference of their reciprocals is $\frac{5}{14}$. Find the numbers. [CBSE 2014]

Answer 1

Let the required natural numbers be x and (x + 5).

Now, x < x + 5
$\therefore \frac{1}{x}>\frac{1}{x+5}$

According to the given condition,
$$\begin{gathered} \frac{1}{x}-\frac{1}{x+5}=\frac{5}{14} \\ \Rightarrow \frac{x+5-x}{x\left(x+5\right)}=\frac{5}{14} \\ \Rightarrow \frac{5}{x^2+5x}=\frac{5}{14} \\ \Rightarrow x^2+5x=14 \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2+5x-14=0 \\ \Rightarrow x^2+7x-2x-14=0 \\ \Rightarrow x\left(x+7\right)-2\left(x+7\right)=0 \\ \Rightarrow \left(x+7\right)\left(x-2\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x+7=0\mathrm{or}x-2=0 \\ \Rightarrow x=-7\mathrm{or}x=2 \end{gathered}$$
∴ x = 2 (−7 is not a natural number)

When x = 2,
x + 5 = 2 + 5 = 7

Hence, the required natural numbers are 2 and 7.