Suppose, first pipe alone takes $ x $ hours to fill the tank .
Then, second and third pipes will take$\left ( x -5\right)$ and $\left( x - 9\right)$ hours respectively to fill the tank.
$\therefore \dfrac{1}{x} $+$ \dfrac{1}{(x - 5)} $=$ \dfrac{1}{(x - 9)} $
$\Rightarrow \dfrac{x - 5 + x}{x(x - 5)} $=$ \dfrac{1}{(x - 9)} $
$\Rightarrow$ $\left(2 x - 5\right)$$\left( x - 9\right)$ = $ x $$\left( x - 5\right)$
$\Rightarrow x $2 - 18$ x $ + 45 = 0
$\left( x - 15\right)$$\left( x - 3\right)$ = 0
$\Rightarrow x $ = 15. [neglecting $ x $ = 3]
Part filled in 2 hours =$ \dfrac{2}{6} $=$ \dfrac{1}{3} $
Remaining part =$ \left(1 -\dfrac{1}{3} \right) $=$ \dfrac{2}{3} $.
$\therefore$ $\left(A + B\right)$s 7 hours work =$ \dfrac{2}{3} $
$\left(A + B\right)$s 1 hours work =$ \dfrac{2}{21} $
$\therefore$ Cs 1 hours work = { (A + B + C)s 1 hours work } - { (A + B)s 1 hours work }
=$ \left(\dfrac{1}{6} -\dfrac{2}{21} \right) $=$ \dfrac{1}{14} $
$\therefore$ C alone can fill the tank in 14 hours.
Part filled in 4 minutes = 4$ \left(\dfrac{1}{15} +\dfrac{1}{20} \right) $=$ \dfrac{7}{15} $.
Remaining part =$ \left(1 -\dfrac{7}{15} \right) $=$ \dfrac{8}{15} $.
Part filled by B in 1 minute =$ \dfrac{1}{20} $
$\therefore \dfrac{1}{20} $:$ \dfrac{8}{15} $:: 1 : $ x $
$ x $ =$ \left(\dfrac{8}{15} \times 1 \times 20\right) $= 10$ \dfrac{2}{3} $min = 10 min. 40 sec.
$\therefore$ The tank will be full in $\left(4 min. + 10 min. + 40 sec.\right)$ = 14 min. 40 sec.
$\left(A + B\right)$s 1 hours work =$ \left(\dfrac{1}{12} +\dfrac{1}{15} \right) $=$ \dfrac{9}{60} $=$ \dfrac{3}{20} $.
$\left(A + C\right)$s hours work =$ \left(\dfrac{1}{12} +\dfrac{1}{20} \right) $=$ \dfrac{8}{60} $=$ \dfrac{2}{15} $.
Part filled in 2 hrs =$ \left(\dfrac{3}{20} +\dfrac{2}{15} \right) $=$ \dfrac{17}{60} $.
Part filled in 6 hrs =$ \left(3 \times\dfrac{17}{60} \right) $=$ \dfrac{17}{20} $.
Remaining part =$ \left(1 -\dfrac{17}{20} \right) $=$ \dfrac{3}{20} $.
Now, it is the turn of A and B and $ \dfrac{3}{20} $ part is filled by A and B in 1 hour.
$\therefore$ Total time taken to fill the tank = $\left(6 + 1\right)$ hrs = 7 hrs.
Part filled by $\left(A + B + C\right)$ in 3 minutes = 3$\left(\dfrac{1}{30} +\dfrac{1}{20} +\dfrac{1}{10} \right) $=$ \left(3 \times\dfrac{11}{60} \right) $=$ \dfrac{11}{20} $.
Part filled by C in 3 minutes =$ \dfrac{3}{10} $.
$\therefore$ Required ratio =$ \left(\dfrac{3}{10} \times\dfrac{20}{11} \right) $=$ \dfrac{6}{11} $.
Part filled by pipe A in 1 hour $=\dfrac{1}{5}$
Part filled by pipe B in 1 hour =$\dfrac{1}{20}$
Part filled by pipe A and B in 1 hour $=\dfrac{1}{5}+\dfrac{1}{20}$=$\dfrac{1}{4}$
i.e., A and B together can fill the tank in 4 hours.
Due to leakage, it took 40 minutes more to fill the tank.
i.e., due to the leakage, the tank got filled in
4$\dfrac{40}{60}$ hour =4$\dfrac{2}{3}$ hour =$\dfrac{14}{3}$ hour.
Net part filled by pipe A, pipe B and the leak in 1 hour=$\dfrac{3}{14}$
Therefore, part emptied by the leak in 1 hour=$\dfrac{1}{4}-\dfrac{3}{14}=\dfrac{1}{28}$
i.e., the leak can empty the tank in 28 hours.
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\dfrac{1}{6} \right) $=$ \dfrac{2}{3} $.
Remaining part =$ \left(1 -\dfrac{1}{2} \right) $=$ \dfrac{1}{2} $.
$\therefore \dfrac{2}{3} $:$ \dfrac{1}{2} $:: 1 : $ x $
$\Rightarrow x $ =$ \left(\dfrac{1}{2} \times 1 \times\dfrac{3}{2} \right) $=$ \dfrac{3}{4} $hours i.e., 45 mins.
So, total time taken = 3 hrs. 45 mins.
Since pipe B is faster than pipe A, the tank will be emptied.
Part filled by pipe A in 1 minute $=\dfrac{1}{12}$
Part emptied by pipe B in 1 minute $=\dfrac{1}{6}$
Net part emptied by pipe A and B in 1 minute
$=\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}$
Time taken to empty $\dfrac{2}{5}$ of the tank
$=\dfrac{\left(\dfrac{2}{5}\right)}{\left(\dfrac{1}{12}\right)}$=$\dfrac{2}{5}$×12=4.8 min
Let B be turned off after $ x $ minutes. Then,
Part filled by $\left(A + B\right)$ in $ x $ min. + Part filled by A in $\left(30 - x \right)$ min. = 1.
$\therefore x \left(\dfrac{2}{75} +\dfrac{1}{45} \right) $+ $\left(30 - x \right)$.$ \dfrac{2}{75} $= 1
$\Rightarrow \dfrac{11x}{225} $+$ \dfrac{(60 -2x)}{75} $= 1
$\Rightarrow$ 11$ x $ + 180 - 6$ x $ = 225.
$\Rightarrow x $ = 9.
Let the cistern be filled by pipe A alone in $ x $ hours.
Then, pipe B will fill it in $\left( x + 6\right)$ hours.
$\therefore \dfrac{1}{x} $+$ \dfrac{1}{(x + 6)} $=$ \dfrac{1}{4} $
$\Rightarrow \dfrac{x + 6 + x}{x(x + 6)} $=$ \dfrac{1}{4} $
$\Rightarrow x $2 - 2$ x $ - 24 = 0
$\Rightarrow$ $\left( x -6\right)$$\left( x + 4\right)$ = 0
$\Rightarrow x $ = 6. [neglecting the negative value of $ x $]