Question

The difference of two natural numbers is 3 and the difference of their reciprocals is $\frac{3}{28}$. Find the numbers. [CBSE 2014]

Answer 1

Let the required natural numbers be x and (x + 3).

Now, x < x + 3
$\therefore \frac{1}{x}>\frac{1}{x+3}$

According to the given condition,
$$\begin{gathered} \frac{1}{x}-\frac{1}{x+3}=\frac{3}{28} \\ \Rightarrow \frac{x+3-x}{x\left(x+3\right)}=\frac{3}{28} \\ \Rightarrow \frac{3}{x^2+3x}=\frac{3}{28} \\ \Rightarrow x^2+3x=28 \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2+3x-28=0 \\ \Rightarrow x^2+7x-4x-28=0 \\ \Rightarrow x\left(x+7\right)-4\left(x+7\right)=0 \\ \Rightarrow \left(x+7\right)\left(x-4\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x+7=0\mathrm{or}x-4=0 \\ \Rightarrow x=-7\mathrm{or}x=4 \end{gathered}$$
∴ x = 4 (−7 is not a natural number)

When x = 4,
x + 3 = 4 + 3 = 7

Hence, the required natural numbers are 4 and 7.