It is given that two pipes can fill a tank in 12 minutes and 20 minutes respectively.
Tank filled by first pipe in 1 minute =1/12
Tank filled by second pipe in 1 minute =1/20
Tank filled by both pipes in 1 minute = 1/12 + 1/20 =1/10
Let first pipe was open for (10 -x) minutes.
$\dfrac{1}{12} \times (10-x) + \dfrac{1}{20} \times 10 $=1
$\dfrac{10}{12}-\dfrac{x}{12}+\dfrac{1}{2} $ =1
$\dfrac{16}{12} -1 =\dfrac{x}{12} $
$\dfrac{4}{12} =\dfrac{x}{12}$
x=4
Since the first pipe was open for (10-x) minutes,
therefore first pipe was open for 6 minutes.
Part of tank filled by all three pipes in one minute
= 1/15 + 1/12 – 1/20 = $\dfrac{8+10-6}{120}$ =$\dfrac{18-6}{120}
=1/10
So, the tank becomes full in 10 minutes.
Let leak can empty the full tank in x hours.
=> 1/12 - 1/x = 1/20 => $\dfrac{1}{12}$ -$\dfrac{1}{20} $ =$\dfrac{1}{x}$
=>$\dfrac{5-3}{60}
=1/30
x=30 hours
So, the leak alone can empty the full tank in 30 hours.
Part filled by first tap in 1 hour =$\dfrac{1}{3}$
Part emptied by second tap 1 hour $=\dfrac{1}{8}$
Net part filled by both these taps in 1 hour
=$\dfrac{1}{3}-\dfrac{1}{8}=\dfrac{5}{24}$
i.e, the cistern gets filled in $\dfrac{24}{5}$hours =4.8 hours.
Work done by the waste pipe in 1 minute =$ \dfrac{1}{15} $-$ \left(\dfrac{1}{20} +\dfrac{1}{24} \right) $
=$ \left(\dfrac{1}{15} -\dfrac{11}{120} \right) $
= - $ \dfrac{1}{40} $. [-ve sign means emptying]
$\therefore$ Volume of$ \dfrac{1}{40} $part = 3 gallons.
Volume of whole = $\left(3 \times 40\right)$ gallons = 120 gallons.
Work done by the leak in 1 hour =$ \left(\dfrac{1}{2} -\dfrac{3}{7} \right) $=$ \dfrac{1}{14} $.
$\therefore$ Leak will empty the tank in 14 hrs.
Part filled by pipe A in 1 hour =$\dfrac{1}{8}$
Part filled by pipe B in 1 hour =$\dfrac{1}{4}$
Part filled by pipe C in 1 hour =$\dfrac{1}{24}$
Part filled by pipes A,B,C together in 1 hour=$\dfrac{1}{8}+\dfrac{1}{4}+\dfrac{1}{24}=\dfrac{10}{24}$
i.e, pipes A,B,C together can fill the tank in $\dfrac{24}{10}$ hour =2.4 hour.
Part filled by pipe A in 1 hr $=\dfrac{1}{8}$
Part filled by pipe B in 1 hr $=\dfrac{1}{24}$
Part emptied by pipe C in 1 hr $=\dfrac{1}{12}$
Net part filled by pipe A, B, C together in $1$ hr
$=\dfrac{1}{8}+\dfrac{1}{24}-\dfrac{1}{12}=\dfrac{2}{24}=\dfrac{1}{12}$
i.e, the tank will be filled in $12$ hr.
Part filled by first pipe in $1$ minute $=\dfrac{1}{2}$
Part filled by second pipe in $1$ minute $=\dfrac{1}{6}$
Net part filled by pipe A and B in $1$ minute
$=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{2}{3}$
i.e, pipe A and B together can fill the tank in $\dfrac{3}{2}$ minutes $=1.5$ minutes.
Part filled by $\left(A + B\right)$ in 1 minute =$ \left(\dfrac{1}{60} +\dfrac{1}{40} \right) $=$ \dfrac{1}{24} $.
Suppose the tank is filled in $ x $ minutes.
Then,$ \dfrac{x}{2} \left(\dfrac{1}{24} +\dfrac{1}{40} \right) $= 1
$\Rightarrow \dfrac{x}{2} $x$ \dfrac{1}{15} $= 1
$\Rightarrow x $ = 30 min.
Capacity of the tank =$(13 \times 51)$ litre.
Number of buckets required when capacity of each bucket is 17 litre
=$\dfrac{13×51}{17}=13×3=39$
Let faster pipe alone can fill the tank in $x$ minutes.
Then, slower pipe alone can fill the tank in $6x$ minutes.
$\dfrac{x×6x}{x+6x}=22\dfrac{6x}{7}=22x=\dfrac{22×7}{6}=\dfrac{154}{6}$
Time required for the slower pipe to fill the tank
=$6x=154$ minute.
Suppose pipe A alone takes $ x $ hours to fill the tank.
Then, pipes B and C will take$ \dfrac{x}{2} $and$ \dfrac{x}{4} $hours respectively to fill the tank.
$\therefore \dfrac{1}{x} $+$ \dfrac{2}{x} $+$ \dfrac{4}{x} $=$ \dfrac{1}{5} $
$\Rightarrow \dfrac{7}{x} $=$ \dfrac{1}{5} $
$\Rightarrow x $ = 35 hrs.