Two taps $A$ and $B$ can fill a tank in $20$ minutes and $30$ minutes respectively. An outlet pipe $C$ can empty the full tank in $15$ minutes. $A$,$B$ and $C$ are opened alternatively each for $1$ minute. How long will the tank take to be filled $?$
Below is my approach :-
For 1$^{st}$ set of $3$ minutes ; net work done by the opening of two inlet taps and one outlet tap = $\frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{1}{60}$
For the $59$ sets of 3 minutes ; work done = $\frac{59}{60}$ $\Rightarrow$ Total time till now = $59\times3=177$ minutes
Work left = $\frac{1}{60}$
Now after $177$ minutes, again tap $A$ will be opened and as we can see that tap $A$ can do $\frac{1}{20}$ units of work in $1$ minute and the units of work left is pretty much smaller quantity i.e. $\frac{1}{60}$ units, then the whole tank will be filled under a minute only by the tap $A$ which means the answer to this problem should lie between $177$ and $178$ minutes but to my bad luck the correct answer has been given as $167$ minutes. Can someone please explain me where I am going wrong?
Your thought that instead of being full at the end of the minute C discharges, which would give $180$ minutes, you should look at shorter times when the tank gets full is a good one but you didn't push it far enough. After some number $n$ of $3$ minute intervals the tank is $\frac n{60}$ full. In the next $2$ minutes the $A$ tap fills $\frac 1{20}$ and then $B$ fills $\frac 1{30}$. We want the tank full at this moment. We have $$\frac n{60}+\frac 1{20}+\frac 1{30}=1\\ \frac{n+5}{60}=1\\ n=55$$ The three minute intervals take $3 \cdot 55=165$ minutes, then there are the two minutes for the $A$ and $B$ taps making $167$ in total.