Two pipes running together can fill an empty cistern in $4\dfrac{20}{21}$ minutes. If one pipe takes 5 minutes more than the other pipe to fill the empty cistern. Find the time in which each pipe would fill the cistern.

Let the time taken by each pipe be x minutes and y minutes.

It is given that one pipe takes 5 minutes more than the other pipe to fill the empty cistern.

⇒ x = y + 5

So, one minute work of the pipe = $\dfrac{1}{x} = \dfrac{1}{y + 5}$

And, one minute work of the other pipe = $\dfrac{1}{y}$

Both pipes running together can fill the empty cistern in $4\dfrac{20}{21} = \dfrac{104}{21}$ minutes.

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{1}{y + 5} + \dfrac{1}{y} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{1 \times y}{y(y + 5)} + \dfrac{1 \times (y + 5)}{y(y + 5)} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{y}{y(y + 5)} + \dfrac{y + 5}{y(y + 5)} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{y + y + 5}{y(y + 5)} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{2y + 5}{y^2 + 5y} = \dfrac{21}{104}\\[1em] \Rightarrow 104(2y + 5) = 21(y^2 + 5y)\\[1em] \Rightarrow 208y + 520 = 21y^2 + 105y\\[1em] \Rightarrow 21y^2 + 105y - 208y - 520 = 0\\[1em] \Rightarrow 21y^2 - 103y - 520 = 0\\[1em] \Rightarrow 21y^2 + 65y - 168y - 105 = 0\\[1em] \Rightarrow y(21y + 65) - 8(21y + 65) = 0\\[1em] \Rightarrow (21y + 65)(y - 8) = 0\\[1em] \Rightarrow (21y + 65) = 0 \text{ or }(y - 8) = 0\\[1em] \Rightarrow y = -\dfrac{65}{21} \text{ or }y = 8\\[1em]$

As minutes cannot be negative.

So, one pipe takes 8 minutes and the other pipe takes 8 + 5 = 13 minutes.

Hence, the time taken by pipes = 8 minutes and 13 minutes.