The difference of two natural numbers is 5 and sum of their reciprocals is 3/10 $\dfrac{3}{10}$. Find the two numbers.

Let the two natural numbers be: x and x + 5

Given,

The sum of the reciprocals of two numbers is $\dfrac{3}{10}$.

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{x + 5} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{x + 5 + x}{x(x + 5)} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2x + 5}{x(x + 5)} = \dfrac{3}{10} \\[1em] \Rightarrow 10(2x + 5) = 3[x(x + 5)] \\[1em] \Rightarrow 20x + 50 = 3x^2 + 15x \\[1em] \Rightarrow 3x^2 + 15x - 20x - 50 = 0 \\[1em] \Rightarrow 3x^2 - 5x - 50 = 0 \\[1em] \Rightarrow 3x^2 + 10x - 15x - 50 = 0 \\[1em] \Rightarrow x(3x + 10) - 5(3x + 10) = 0 \\[1em] \Rightarrow (x - 5)(3x + 10) = 0 \\[1em] \Rightarrow (x - 5) = 0 \text{ or } (3x + 10) = 0 \\[1em] \Rightarrow x = 5 \text{ or } 3x = -10 \\[1em] \Rightarrow x = 5 \text{ or } x = -\dfrac{10}{3}$

Since, the numbers are natural numbers, thus x cannot be negative.

Thus, x = 5 and x + 5 = 5 + 5 = 10.

Hence, numbers are 5 and 10.