3 pipes at a time #1

First, we have to $a)$ find out how full the cistern will be in 5 minutes, then $b)$ find the net water loss per minute when pipes A, B and C are kept open. Then, $c)$ we calculate how many minutes it will take to empty the cistern filled with the amount of water found in $a)$ using the net water loss rate from $b)$ .

$a)$ Pipe A can fill up $\frac{1}{12}$ of the cistern in a minute, and pipe B can fill up $\frac{1}{15}$ of the cistern in a minute. Therefore, in five minutes, Pipes A and B can fill up $\frac{5}{12}$ and $\frac{5}{15}$ of the pipe respectively.

$\dfrac{5}{15} = \dfrac{1}{3} = \dfrac{4}{12}$

$\dfrac{5}{12} + \dfrac{4}{12} = \dfrac{9}{12} = \dfrac{3}{4}$

$\frac{3}{4}$ of the cistern can be filled up by pipes A and B in five minutes.

$b)$ In a minute, pipes A and B can fill up $\frac{1}{12}$ and $\frac{1}{15}$ of the cistern respectively, and pipe C can drain $\frac{1}{6}$ of the cistern.

$\dfrac{1}{12} + \dfrac{1}{15} = \dfrac{5}{60} + \dfrac{4}{60} = \dfrac{9}{60}$

$\dfrac{9}{60} - \dfrac{1}{6} = \dfrac{9}{60} - \dfrac{10}{60} = -\dfrac{1}{60}$

The net water loss rate is $\frac{1}{60}$ of the cistern per minute.

$c)$ Since the new water loss rate per minute is $\frac{1}{60}$ , the cistern can be emptied from a full tank in 60 minutes with pipes A, B and C kept open. After five minutes, the cistern is only $\frac{3}{4}$ filled, so it will only take $\frac{3}{4}$ of the time to drain it. $\frac{3}{4}$ of 60 is 45. Hence, it will take $\boxed{45}$ minutes.