One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.

Let second pipe fill in x hours, and first pipe in (x - 3) hours.

In one hour second pipe will fill $\dfrac{1}{x}$ and first pipe $\dfrac{1}{x - 3}$.

Given, together pipes can fill the cistern in 6 hours 40 minutes i.e. $\dfrac{400}{60}$ hours.

Hence, in one hour they will fill $\dfrac{1}{\dfrac{400}{60}} = \dfrac{60}{400}$.

$\therefore \dfrac{1}{x} + \dfrac{1}{x - 3} = \dfrac{60}{400} \\[1em] \Rightarrow \dfrac{x - 3 + x}{x(x - 3)} = \dfrac{3}{20} \\[1em] \Rightarrow \dfrac{2x - 3}{x^2 - 3x} = \dfrac{3}{20} \\[1em] \Rightarrow 20(2x - 3) = 3(x^2 - 3x) \\[1em] \Rightarrow 40x - 60 = 3x^2 - 9x \\[1em] \Rightarrow 3x^2 - 9x - 40x - 60 = 0 \\[1em] \Rightarrow 3x^2 - 49x - 60 = 0 \\[1em] \Rightarrow 3x^2 - 45x - 4x - 60 = 0 \\[1em] \Rightarrow 3x(x - 15) - 4(x - 15) = 0 \\[1em] \Rightarrow (3x - 4)(x - 15) = 0 \\[1em] \Rightarrow 3x - 4 = 0 \text{ or } x - 15 = 0 \\[1em] \Rightarrow x = \dfrac{4}{3} \text{ or } x = 15.$

Since first pipe takes 3 hours less than second pipe,

∴ x ≠ $\dfrac{4}{3}$ as in this case time will be negative which is not possible.

∴ x = 15, x - 3 = 12.

Hence, pipes will take 12 hours and 15 hours to fill cistern separately.