Let the slower pipe alone fill the tank in x minutes

Then, faster pipe will fill it in \(\frac{3}{x}\) minutes.

So, \(\frac{1}{x}+\frac{3}{x}=\frac{1}{36}\)

   \(\frac{4}{x}=\frac{1}{36}\)

x = 144 min.

Part filled by (A + B) in 1 minute =  \(\left(\frac{1}{60}+\frac{1}{40}\right)=\frac{1}{24}\)

Suppose the tank is filled in x minutes.

Then,  \(\frac{x}{2}\left(\frac{1}{24}+\frac{1}{40}\right)=1\)

  \(\frac{x}{2}\times\frac{1}{15}=1\)

x=30min

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour = \(\left(4\times\frac{1}{6}\right)=\frac{2}{3}\)

Remaining part = \(\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

So,  \(\frac{2}{3}:\frac{1}{2}::1:x\)

x=     \(\left(\frac{1}{2}\times1\times\frac{3}{2}\right)=\frac{3}{4}\)   hours i.e., 45 mins.

So, total time taken = 3 hrs. 45 mins

(A + B)'s 1 hour's work = \(\left(\frac{1}{12}+\frac{1}{15}\right)=\frac{9}{60}=\frac{3}{20}\)

A + C)'s hour's work = \(\left(\frac{1}{12}+\frac{1}{20}\right)=\frac{8}{60}=\frac{2}{15}\)

Part filled in 2 hrs =\(\left(\frac{3}{20}+\frac{2}{15}\right)=\frac{17}{60}\)

Part filled in 6 hrs = \(\left(3\times\frac{17}{60}\right) =\frac{17}{20}\)

Remaining part = \(\left(1-\frac{17}{20}\right) =\frac{3}{20}\)

Now, it is the turn of A and B and \(\frac{3}{20}\)  part is filled by A and B in 1 hour

Part filled in 2 hours = \(\frac{2}{6} =\frac{1}{3}\)

Remaining part = \(\left(1-\frac{1}{3}\right) =\frac{2}{3}\)

So, (A + B)'s 7 hour's work =  \(\frac{2}{3}\)

(A + B)'s 1 hour's work =  \(\frac{2}{21}\)

So, C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=  \(\left(\frac{1}{6}-\frac{2}{21}\right)=\frac{1}{14}\)

So, C alone can fill the tank in 14 hours.