Part filled by A in 1 min =$ \dfrac{1}{20} $.
Part filled by B in 1 min =$ \dfrac{1}{30} $.
Part filled by $\left(A + B\right)$ in 1 min =$ \left(\dfrac{1}{20} +\dfrac{1}{30} \right) $=$ \dfrac{1}{12} $.
$\therefore$ Both pipes can fill the tank in 12 minutes.
Net part filled in 1 hour$ \left(\dfrac{1}{5} +\dfrac{1}{6} -\dfrac{1}{12} \right) $=$ \dfrac{17}{60} $.
$\therefore$ The tank will be full in$ \dfrac{60}{17} $hours i.e., 3$ \dfrac{9}{17} $hours.
Let the slower pipe alone fill the tank in $ x $ minutes.
Then, faster pipe will fill it in$ \dfrac{x}{3} $minutes.
$\therefore \dfrac{1}{x} $+$ \dfrac{3}{x} $=$ \dfrac{1}{36} $
$\Rightarrow \dfrac{4}{x} $=$ \dfrac{1}{36} $
$\Rightarrow x $ = 144 min.
Part filled by pipe A in 1 minute =$\dfrac{1}{40}$
Part filled by pipe B in 1 minute $=\dfrac{1}{30}$
Net part filled by pipe A, pipe B and the third pipe together in 1 minute =$\dfrac{1}{20}$
Therefore, part emptied by the third pipe in 1 minute =$\dfrac{1}{40}+\dfrac{1}{30}-\dfrac{1}{20}=\dfrac{3+4-6}{120}=\dfrac{1}{120}$
i.e., third pipe alone can empty the cistern in 120 minutes.
Part filled by P1 in 1 minute =1/30
Suppose P3 can empty the cistern in x minutes. Then,
Part emptied by P3 in 1 minute =1/x
Net part emptied by all pipes together in 1 min
=$\dfrac{1}{42×60}=\dfrac{1}{2520}$
$\dfrac{1}{x}−\dfrac{1}{30}−\dfrac{1}{20}=\dfrac{1}{2520}$
⇒$\dfrac{1}{x}=\dfrac{211}{2520}$
⇒x=$\dfrac{2520}{211}$
i.e., P3 alone can empty the cistern in $\dfrac{2520}{211}$ minutes.
Capacity of the cistern =$\dfrac{2520}{211}$×11≈131.4 litre
Let inflow alone can fill the tank in x hours.
Then, the leak alone can empty the tank in 2x hours.
Part filled by the inflow in 1 hour =$\dfrac{1}{x}$
Part emptied by the leak in 1 hour =$\dfrac{1}{2x}$
Net part filled by the inflow and the leak in 1 hour =$\dfrac{1}{x+36}$
$\dfrac{1}{x}−\dfrac{1}{2x}=\dfrac{1}{x+36}$
⇒$\dfrac{1}{2x}=\dfrac{1}{x+36}$
⇒2x=x+36
⇒x=36
The faster pipe is equivalent to 3 slower pipes. Both them together is equivalent to 4 slower pipes.
Therefore, 4 slower pipes together can fill the tank in 36 minutes.
=> 1 slower pipe can fill the tank in 36×4 minute.
=> 1 faster pipe can fill the tank in $\dfrac{36×4}{3}$=48 minutes.
The first two pipes together can fill the tank in $\dfrac{20×30}{20+30}$=12 minutes.
Therefore, quantity filled by first pipe and second pipe together in 4 minutes is equal to quantity emptied by waste pipe in 12 minutes.
i.e., quantity filled by first pipe and second pipe together in 12 minutes is equal to quantity emptied by waste pipe in 36 minutes.
i.e., waste pipe can empty the tank in 36 minutes.
60 minute can pipe in liquid equal to $\dfrac{60}{7.5}$=8 times the capacity of the cistern.
Pipe B in 60 minute can pipe in liquid equal to $\dfrac{60}{5}$=12 times the capacity of the cistern.
Tank was full when all the three pipes were opened. So, pipe C emptied liquid equal to 8+12+1=21 times the capacity of the cistern in 1 hour.
i.e., 21 times the capacity of the cistern =14×60
Capacity of the tank =$\dfrac{14×60}{21}$=40 litre
Suppose B alone takes x hours to fill the cistern.
Then, A alone takes (x−40) hours to fill the cistern
$\dfrac{x(x−40)}{x+x−40}$=15
x2−40x=30x−600
x2−70x+600=0
(x−60)(x−10)=0
x=60