Cistern filled by A and B in 5 ninutes.

= 5 $\displaystyle \left(\frac{1}{20} + \frac{1}{25}\right) = \frac{9}{20}$

Part unfilled = $\displaystyle \left(1 - \frac{9}{20}\right) = \frac{11}{20}$

Now $\displaystyle \frac{1}{20}$ part is filled by A in 1 minute.

$\displaystyle \frac{11}{20}$ part wiil be filled by A in 20% of $\displaystyle \frac{11}{20}$ = 11 minutes.

Cistern filled in 5 ninutes.

= 5 $\displaystyle \left(\frac{1}{12} + \frac{1}{15}\right) = \frac{3}{4}$ of tank.

After 5 minutes, let the third pipe emptied be "t" minutes, then = t $\displaystyle \left(\frac{1}{12} + \frac{1}{15}\right) + \frac{3}{4} = \frac{t}{6}$

t = 45 minutes.

Water tank filled in 10 hours.

= 10 $\displaystyle \left(\frac{1}{15} + \frac{1}{20} - \frac{1}{25}\right) = \frac{23}{30}$

Remaining part = 1 - $\displaystyle \frac{23}{30} = \frac{7}{30}$

Now work done by (A + B) in 1 hour = $\displaystyle \left(\frac{1}{15} + \frac{1}{20}\right) = \frac{7}{60}$

It means $\displaystyle \frac{7}{60}$ part is filled in 1 hour.

So, $\displaystyle \frac{7}{30}$ part is filled in = $\displaystyle \frac{60}{7} \times \frac{7}{30}$ = 2 hours.

Cistern filled by (A + B) in 1 minute.

= $\displaystyle \frac{2}{15} + \frac{1}{5} = \frac{1}{3}$

Let the capacity of tank be "Z" litres.

In one hour, Inlet in litres + Z litres = outlet in litres.

= 60 $\displaystyle \times \frac{\text{Z}}{3}$ + Z = 14 x 60 = Z = 40.

Hence, the capacity of cistern = 40 litres.

Let the second pipe be turned of after "t" minutes.

Then, work done by (A + B) in x minutes is => t $\displaystyle \left(\frac{1}{24} + \frac{1}{32}\right) = \frac{7\text{t}}{92} => \left(1\right)$

Also, work done by A in (18-t) minutes = $\displaystyle \frac{18-\text{t}}{24} => \left(2\right)$

Now (1) + (2) => $\displaystyle \frac{7\text{t}}{96} + \frac{18-\text{x}}{24}$ = 1.

t = 8.

Work done in 3 minutes by 3 pipes.

Then, work done by (A + B) in x minutes is = $\displaystyle \frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{1}{60}$

It means $\displaystyle \frac{1}{60}$ of tank is filled in 3 minutes.

$\displaystyle \frac{55}{60}$ part of the cistern is filled in 3 x 55 or 165 minutes.

Out of remaining $\displaystyle \frac{5}{60}$ cistern, $\displaystyle \frac{1}{20}$ is filled by A in 1 minute, and remaining is

$\displaystyle \frac{5}{60} - \frac{1}{20}$ i.e. $\displaystyle \frac{1}{30}$ is filled by B in 1 minute.

Therefore answer is, 165 + 2 = 167 minutes.

Work done by both in 4 minutes.

= 4 $\displaystyle \left(\frac{1}{12} + \frac{1}{16}\right) = \frac{7}{12}$

Remaining part of cistern = $\displaystyle \left(1 - \frac{7}{12}\right) = \frac{5}{12}$

Now $\displaystyle \frac{5}{12}$ part is filled by P in $\displaystyle \frac{5}{12} \times \frac{12}{1}$ = 5 minutes.

$\displaystyle \frac{1}{7}$ of tank water leak in 8 hours.

So, full tank leak in 7 x 8 = 56 hours.