Pipes and Cistern Questions in English

(B) The total time taken by the pipe to fill the tank is $6 \ h$.
Since the pipe fills the entire tank (which represents $1$ whole unit) in $6 \ h$,the part of the tank filled in $1 \ h$ is calculated by dividing the total work by the total time.
Part of the tank filled in $1 \ h = \frac{1}{6}$.

(D) The pipe fills $\frac{1}{8}$ of the tank in $1 \ h$.
To fill the entire tank (which is $1$ whole part),we need to calculate the time required.
Time taken to fill $1$ full tank $= \frac{1 \ h}{1/8} = 1 \times 8 = 8 \ h$.
Therefore,the pipe will take $8 \ h$ to fill the empty tank.

(B) The total time taken to empty the entire cistern is $3$ $h$.
To find the time required to empty $\frac{2}{3}$ of the cistern,we multiply the total time by the fraction of the cistern to be emptied.
Time taken $= 3 \times \frac{2}{3} = 2$ $h$.
Therefore,the pipe will empty two-third of the cistern in $2$ $h$.

(D) Let the total capacity of the tank be $1$ unit.
Rate of tap $A$ and tap $B$ working together $= \frac{1}{40}$ tank per minute.
Rate of tap $A$ working alone $= \frac{1}{60}$ tank per minute.
Let the time taken by tap $B$ alone to fill the tank be $x$ minutes.
Therefore,the rate of tap $B$ is $\frac{1}{x}$ tank per minute.
We know that: $(\text{Rate of } A) + (\text{Rate of } B) = \text{Combined Rate}$.
$\frac{1}{60} + \frac{1}{x} = \frac{1}{40}$.
$\frac{1}{x} = \frac{1}{40} - \frac{1}{60}$.
Taking the least common multiple $(LCM)$ of $40$ and $60,$ which is $120$:
$\frac{1}{x} = \frac{3 - 2}{120} = \frac{1}{120}$.
Thus,$x = 120$ minutes.

(B) Let the capacity of the tank be the least common multiple of $10$ and $6$, which is $30$ units.
Efficiency of the filling pipe $= 30 / 10 = 3$ units per hour.
Efficiency of the emptying pipe $= 30 / 6 = 5$ units per hour.
When both pipes are opened simultaneously, the net rate of emptying the tank $= 5 - 3 = 2$ units per hour.
Time taken to empty the tank $= \text{Total capacity} / \text{Net rate} = 30 / 2 = 15 \, h$.

(B) Let the capacity of the cistern be the $LCM$ of $3, 4,$ and $5$, which is $60 \, \text{units}$.
Efficiency of the first tap $(A) = \frac{60}{3} = 20 \, \text{units/h}$.
Efficiency of the second tap $(B) = \frac{60}{4} = 15 \, \text{units/h}$.
Efficiency of the third tap $(C) = -\frac{60}{5} = -12 \, \text{units/h}$ (since it empties the cistern).
When all three taps are opened simultaneously, the net efficiency $= 20 + 15 - 12 = 23 \, \text{units/h}$.
Time taken to fill the cistern $= \frac{\text{Total Capacity}}{\text{Net Efficiency}} = \frac{60}{23} \, h$.
Converting this to a mixed fraction, we get $2 \frac{14}{23} \, h$.

(A) Let the total capacity of the tank be the Least Common Multiple $(LCM)$ of $30, 10,$ and $40$,which is $120$ units.
Efficiency of pipe $A = 120 / 30 = 4 \text{ units/min}$.
Efficiency of pipe $B = 120 / 10 = 12 \text{ units/min}$.
Efficiency of pipe $C = 120 / 40 = 3 \text{ units/min}$ (emptying,so negative).
When all pipes are opened together,the net efficiency = $4 + 12 - 3 = 13 \text{ units/min}$.
Time required to fill the tank = $\frac{\text{Total Capacity}}{\text{Net Efficiency}} = \frac{120}{13} \text{ minutes}$.
Converting to mixed fraction: $120 / 13 = 9 \frac{3}{13} \text{ minutes}$.

(B) Let the work done by taps $A, B,$ and $C$ in $1$ $min$ be $\frac{1}{a}, \frac{1}{b},$ and $\frac{1}{c}$ respectively.
Given that the three taps together fill the cistern in $10$ $min$,so $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{10}$.
Given that tap $A$ alone fills it in $30$ $min$,so $\frac{1}{a} = \frac{1}{30}$.
Given that tap $B$ alone fills it in $40$ $min$,so $\frac{1}{b} = \frac{1}{40}$.
Substituting these values into the first equation:
$\frac{1}{30} + \frac{1}{40} + \frac{1}{c} = \frac{1}{10}$.
$\frac{1}{c} = \frac{1}{10} - \frac{1}{30} - \frac{1}{40}$.
Taking the least common multiple $(LCM)$ of $10, 30,$ and $40$,which is $120$:
$\frac{1}{c} = \frac{12 - 4 - 3}{120} = \frac{5}{120} = \frac{1}{24}$.
Therefore,tap $C$ alone will take $24$ $min$ to fill the cistern.

(A) Let the capacity of the tank be $300$ units ($LCM$ of $60, 75, 50$).
Efficiency of pipe $A = 300 / 60 = 5$ units/min.
Efficiency of pipe $B = 300 / 75 = 4$ units/min.
Combined efficiency of pipes $A, B,$ and $C = 300 / 50 = 6$ units/min.
Let the efficiency of outlet $C$ be $x$ units/min.
Then, $5 + 4 - x = 6$.
$9 - x = 6 \Rightarrow x = 3$ units/min.
Time taken by $C$ to empty the full tank = $\text{Total Capacity} / \text{Efficiency of } C = 300 / 3 = 100$ minutes.

(D) Let the work done by pipes $A, B$ and $C$ in $1 \text{ h}$ be $\frac{1}{a}, \frac{1}{b}$ and $\frac{1}{c}$ respectively.
Given:
$\frac{1}{a} + \frac{1}{b} = \frac{1}{6}$
$\frac{1}{b} + \frac{1}{c} = \frac{1}{10}$
$\frac{1}{a} + \frac{1}{c} = \frac{1}{12}$
Adding these three equations:
$2 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{1}{6} + \frac{1}{10} + \frac{1}{12}$
$2 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{10 + 6 + 5}{60} = \frac{21}{60} = \frac{7}{20}$
Therefore,$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{7}{40}$
The time taken by $A, B$ and $C$ together to fill the tank is the reciprocal of the combined work rate:
Time $= \frac{40}{7} \text{ h} = 5 \frac{5}{7} \text{ h}$.

(B) Let the efficiency of inlet $B$ be $1$ unit/min. Since inlet $A$ is four times faster,its efficiency is $4$ units/min.
Given that inlet $A$ fills the tank in $15$ minutes,the total capacity of the tank is $4 \times 15 = 60$ units.
When both pipes $A$ and $B$ are opened together,their combined efficiency is $4 + 1 = 5$ units/min.
The time taken to fill the tank together is $\frac{\text{Total Capacity}}{\text{Combined Efficiency}} = \frac{60}{5} = 12$ minutes.

(C) Efficiency of pipe $A = \frac{1}{16}$ tank per hour.
Efficiency of pipe $B = \frac{1}{10}$ tank per hour.
In a cycle of $2 \text{ h}$ (first $A$ for $1 \text{ h}$,then $B$ for $1 \text{ h}$),the part of the tank filled is $\frac{1}{16} + \frac{1}{10} = \frac{5+8}{80} = \frac{13}{80}$.
In $6$ such cycles $(12 \text{ h})$,the part filled is $6 \times \frac{13}{80} = \frac{78}{80} = \frac{39}{40}$.
Remaining part of the tank $= 1 - \frac{39}{40} = \frac{1}{40}$.
Now,it is the turn of pipe $A$ to fill the remaining part.
Time taken by $A$ to fill $\frac{1}{40}$ part $= \frac{1/40}{1/16} = \frac{16}{40} = \frac{2}{5} \text{ h}$.
Total time taken $= 12 \text{ h} + \frac{2}{5} \text{ h} = 12 \frac{2}{5} \text{ h}$.

(D) The pipe takes $27$ $hours$ to empty the full cistern (i.e.,$1$ part).
To find the time required to empty $\frac{2}{3}$ of the cistern,we multiply the total time by the required fraction.
Time taken $= 27 \times \frac{2}{3} \text{ hours}$.
Time taken $= 9 \times 2 = 18 \text{ hours}$.
Therefore,the pipe will empty $\frac{2}{3}$ of the cistern in $18$ $hours$.

(D) Let the total capacity of the tank be $1$ unit.
Rate of pipe $A$ (filling) $= \frac{1}{10}$ units/min.
Rate of pipe $B$ (emptying) $= \frac{1}{6}$ units/min.
Since the rate of emptying is greater than the rate of filling,the tank will be emptied.
Net rate of emptying $= \frac{1}{6} - \frac{1}{10} = \frac{5 - 3}{30} = \frac{2}{30} = \frac{1}{15}$ units/min.
This means the tank is emptied at a rate of $\frac{1}{15}$ of the total capacity per minute.
To empty the $\frac{2}{3}$ part of the tank,the time required is $\text{Time} = \frac{\text{Volume to be emptied}}{\text{Net rate}} = \frac{2/3}{1/15} = \frac{2}{3} \times 15 = 10$ minutes.

(C) The rate of the first tap filling the cistern is $\frac{1}{8}$ of the tank per hour.
The rate of the second tap emptying the cistern is $\frac{1}{16}$ of the tank per hour.
When both taps are opened simultaneously,the net rate of filling is $\frac{1}{8} - \frac{1}{16} = \frac{2-1}{16} = \frac{1}{16}$ of the tank per hour.
Therefore,the time required to fill the tank is the reciprocal of the net rate,which is $16$ $hours$.

(B) Let the capacity of the tank be the least common multiple of $15$ and $10$, which is $30$ units.
Efficiency of the first pipe $= 30 / 15 = 2$ units per hour.
Efficiency of the second pipe $= 30 / 10 = 3$ units per hour.
Combined efficiency of both pipes $= 2 + 3 = 5$ units per hour.
Time taken to empty the full tank $= \text{Total capacity} / \text{Combined efficiency} = 30 / 5 = 6$ $hrs$.

(C) Pipe $A$ fills the cistern in $20$ $minutes$,so its efficiency is $\frac{1}{20}$ part per $minute$.
Pipe $B$ fills the cistern in $25$ $minutes$,so its efficiency is $\frac{1}{25}$ part per $minute$.
When both are opened together,the combined efficiency is $\frac{1}{20} + \frac{1}{25} = \frac{5+4}{100} = \frac{9}{100}$ part per $minute$.
In $5$ $minutes$,the amount filled is $5 \times \frac{9}{100} = \frac{45}{100} = \frac{9}{20}$ part.
The remaining part to be filled is $1 - \frac{9}{20} = \frac{11}{20}$ part.
Since pipe $B$ is turned off,pipe $A$ fills the remaining $\frac{11}{20}$ part.
Time taken by pipe $A = \frac{\text{Remaining part}}{\text{Efficiency of } A} = \frac{11/20}{1/20} = 11$ $minutes$.

(C) Using the work-time formula: $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$
Given:
$M_1 = 12$ (number of pumps)
$D_1 = 15$ (days)
$H_1 = 6$ (hours per day)
$W_1 = 1$ (the reservoir)
For the second case:
$M_2 = x$ (number of pumps)
$D_2 = 12$ (days)
$H_2 = 9$ (hours per day)
$W_2 = 1$ (the same reservoir)
Substituting the values into the formula:
$12 \times 15 \times 6 = x \times 12 \times 9$
Solving for $x$:
$x = \frac{12 \times 15 \times 6}{12 \times 9}$
$x = \frac{15 \times 6}{9}$
$x = \frac{90}{9} = 10$
Therefore,$10$ pumps are required.

(C) Let the capacity of the tank be $C$ litres.
Rate of the leak $= \frac{C}{8}$ litres per hour.
Rate of the tap $= 6 \times 60 = 360$ litres per hour.
When both are open,the net rate of emptying is $\frac{C}{12}$ litres per hour.
The equation is: $\frac{C}{8} - 360 = \frac{C}{12}$.
Rearranging the terms: $\frac{C}{8} - \frac{C}{12} = 360$.
$\frac{3C - 2C}{24} = 360$.
$\frac{C}{24} = 360$.
$C = 360 \times 24 = 8640$ litres.
Therefore,the tank holds $8640$ litres.

(A) Let the capacity of the cistern be $1$ unit.
Rate of the first tap (filling) $= \frac{1}{8}$ units per hour.
Rate of the second tap (emptying) $= \frac{1}{16}$ units per hour.
When both taps are opened simultaneously,the net rate of filling $= \frac{1}{8} - \frac{1}{16}$ units per hour.
Net rate $= \frac{2-1}{16} = \frac{1}{16}$ units per hour.
Time taken to fill the cistern $= \frac{\text{Total capacity}}{\text{Net rate}} = \frac{1}{1/16} = 16$ hours.

(C) Let the capacity of the tank be the Least Common Multiple $(LCM)$ of $5$ and $4$,which is $20$ units.
Efficiency of the filling pipe $(A)$ $= 20 / 5 = 4$ units per hour.
Efficiency of the emptying pipe $(B)$ $= 20 / 4 = 5$ units per hour.
When both pipes are opened,the net change in the tank per hour $= 4 - 5 = -1$ unit per hour.
The negative sign indicates that the tank is being emptied.
Since the tank is already full ($20$ units),the time taken to empty it $= 20 / 1 = 20$ hours.

(A) Let the total capacity of the tank be the Least Common Multiple $(LCM)$ of $15, 12,$ and $20$,which is $60$ units.
Efficiency of the first pipe $= 60 / 15 = 4$ units/$hour$.
Efficiency of the second pipe $= 60 / 12 = 5$ units/$hour$.
Efficiency of the third pipe (emptying) $= - (60 / 20) = -3$ units/$hour$.
When all three pipes are opened together,the net efficiency $= 4 + 5 - 3 = 6$ units/$hour$.
Time taken to fill the tank $= \text{Total Capacity} / \text{Net Efficiency} = 60 / 6 = 10$ $hours$.

(B) Let the capacity of the tank be the $LCM$ of $60$ and $70$,which is $420$ units.
Efficiency of pipe $A = 420 / 60 = 7$ units/min.
Efficiency of pipe $B = 420 / 70 = 6$ units/min.
Let the efficiency of the third pipe (emptying pipe) be $x$ units/min.
When all three pipes are open,the net efficiency = $7 + 6 - x = 13 - x$ units/min.
The tank is filled in $60$ min,so the net efficiency = $420 / 60 = 7$ units/min.
Equating the two: $13 - x = 7$,which gives $x = 6$ units/min.
Time taken by the third pipe to empty the full tank = Total Capacity / Efficiency of third pipe = $420 / 6 = 70$ min.

(D) Pipe $A$ fills the tank in $37 \frac{1}{2} = 75/2$ minutes. Therefore,the rate of pipe $A$ is $2/75$ tank per minute.
Pipe $B$ fills the tank in $45$ minutes. Therefore,the rate of pipe $B$ is $1/45$ tank per minute.
Let pipe $B$ be closed after $x$ minutes. Pipe $A$ remains open for the entire $30$ minutes.
The total work done by pipe $A$ and pipe $B$ must equal $1$ (the full tank).
Work done by $A$ in $30$ minutes + Work done by $B$ in $x$ minutes = $1$.
$30 \times (2/75) + x \times (1/45) = 1$.
$(60/75) + x/45 = 1$.
$4/5 + x/45 = 1$.
$x/45 = 1 - 4/5 = 1/5$.
$x = 45/5 = 9$ minutes.
Thus,pipe $B$ should be closed after $9$ minutes.

(B) Let the total time required to fill the tank be $x$ minutes.
The rate of tap $A$ is $\frac{1}{60}$ tank per minute.
The rate of tap $B$ is $\frac{1}{40}$ tank per minute.
The combined rate of tap $A$ and $B$ is $\frac{1}{60} + \frac{1}{40} = \frac{2+3}{120} = \frac{5}{120} = \frac{1}{24}$ tank per minute.
According to the problem,tap $B$ is used for $\frac{x}{2}$ minutes and both taps $A$ and $B$ are used for the remaining $\frac{x}{2}$ minutes.
Work done by tap $B$ + Work done by $(A+B) = 1$ (full tank).
$\frac{x}{2} \times \frac{1}{40} + \frac{x}{2} \times \frac{1}{24} = 1$
$\frac{x}{80} + \frac{x}{48} = 1$
Taking the $LCM$ of $80$ and $48$,which is $240$:
$\frac{3x + 5x}{240} = 1$
$\frac{8x}{240} = 1$
$\frac{x}{30} = 1$
$x = 30$ minutes.

(A) Let the capacity of the tank be the $LCM$ of $15$ and $18$,which is $90$ units.
Efficiency of tap $A = 90 / 15 = 6$ units/min.
Efficiency of tap $B = 90 / 18 = 5$ units/min.
Combined efficiency of $A$ and $B = 6 + 5 = 11$ units/min.
In $6$ min,taps $A$ and $B$ fill $11 \times 6 = 66$ units.
Let the efficiency of the third tap (emptying) be $E$ units/min.
After the third tap is opened,the net rate of filling/emptying is $(11 - E)$ units/min.
The tank is emptied in $16.5$ min after the third tap is opened,meaning the $66$ units already filled are emptied at a rate of $(E - 11)$ units/min.
$66 = (E - 11) \times 16.5$
$E - 11 = 66 / 16.5 = 4$
$E = 15$ units/min.
Time taken by the third tap to empty the full tank $= 90 / 15 = 6$ min.

(C) Let the total capacity of the tank be $1$ unit.
Combined efficiency of pipes $A, B$, and $C$ is $\frac{1}{6}$ tank per hour.
In $2$ hours, the part of the tank filled by $A, B$, and $C$ is $2 \times \frac{1}{6} = \frac{1}{3}$.
Remaining part of the tank $= 1 - \frac{1}{3} = \frac{2}{3}$.
This remaining $\frac{2}{3}$ part is filled by $A$ and $B$ in $7$ hours.
Therefore, the combined efficiency of $A$ and $B$ is $\frac{2/3}{7} = \frac{2}{21}$ tank per hour.
Efficiency of pipe $C = (\text{Efficiency of } A+B+C) - (\text{Efficiency of } A+B)$.
Efficiency of $C = \frac{1}{6} - \frac{2}{21} = \frac{7 - 4}{42} = \frac{3}{42} = \frac{1}{14}$ tank per hour.
Thus, pipe $C$ alone can fill the tank in $14$ hours.

(D) Let the total capacity of the tank be the Least Common Multiple $(LCM)$ of $4, 6,$ and $3$,which is $12$ units.
Efficiency of tap $A = 12 / 4 = 3$ units/hour.
Efficiency of tap $B = 12 / 6 = 2$ units/hour.
Efficiency of tap $C = -12 / 3 = -4$ units/hour (since it empties the tank).
When all three taps are opened simultaneously,the net efficiency = $3 + 2 - 4 = 1$ unit/hour.
Time taken to fill the tank = Total capacity / Net efficiency = $12 / 1 = 12$ hours.

(B) Let the total capacity of the tank be the least common multiple of $6$ and $5$,which is $30$ units.
Efficiency of tap $A$ (filling) $= 30 / 6 = 5$ units/hour.
Efficiency of tap $B$ (emptying) $= 30 / 5 = 6$ units/hour.
When both taps are opened simultaneously,the net rate of flow $= 5 - 6 = -1$ unit/hour.
The negative sign indicates that the tank is being emptied.
Since the net rate is $1$ unit/hour (emptying),the time required to empty a full tank is $30 / 1 = 30$ hours.

(C) The relationship between the number of pumps $(P)$,time $(T)$,and capacity $(C)$ is given by the formula: $\frac{P_1 \times T_1}{C_1} = \frac{P_2 \times T_2}{C_2}$.
Given:
$P_1 = 18, T_1 = 5 \text{ hours}, C_1 = 1440 \text{ kilo liter}$.
$P_2 = 8, C_2 = 1920 \text{ kilo liter}, T_2 = x$.
Substituting the values into the formula:
$\frac{18 \times 5}{1440} = \frac{8 \times x}{1920}$.
Solving for $x$:
$x = \frac{18 \times 5 \times 1920}{1440 \times 8}$.
$x = \frac{90 \times 1920}{11520}$.
$x = \frac{172800}{11520} = 15 \text{ hours}$.

(D) Part of the cistern filled by two pipes in $1 \text{ h} = \frac{1}{14} + \frac{1}{16} = \frac{8 + 7}{112} = \frac{15}{112}$.
Time taken to fill the cistern without leakage = $\frac{112}{15} \text{ h} = \frac{112}{15} \times 60 \text{ min} = 448 \text{ min}$.
Due to leakage,the time taken to fill the cistern = $448 + 92 = 540 \text{ min} = 9 \text{ h}$.
Let the leakage empty the cistern in $x \text{ h}$.
Net work done in $1 \text{ h} = \frac{1}{9}$.
So,$\frac{15}{112} - \frac{1}{x} = \frac{1}{9}$.
$\frac{1}{x} = \frac{15}{112} - \frac{1}{9} = \frac{135 - 112}{1008} = \frac{23}{1008}$.
$x = \frac{1008}{23} = 43 \frac{19}{23} \text{ h}$.

(C) Efficiency of pipe $P = \frac{1}{12}$ tank per minute.
Efficiency of pipe $R = \frac{1}{15}$ tank per minute.
Efficiency of pipe $M = -\frac{1}{6}$ tank per minute (emptying).
Work done by $P$ and $R$ in $5$ minutes $= 5 \times (\frac{1}{12} + \frac{1}{15}) = 5 \times (\frac{5+4}{60}) = 5 \times \frac{9}{60} = \frac{45}{60} = \frac{3}{4}$ of the tank.
When all three pipes are open,the net efficiency $= \frac{1}{12} + \frac{1}{15} - \frac{1}{6} = \frac{5+4-10}{60} = -\frac{1}{60}$ tank per minute.
The negative sign indicates that the tank is being emptied at a rate of $\frac{1}{60}$ tank per minute.
Time taken to empty $\frac{3}{4}$ of the tank $= \frac{3/4}{1/60} = \frac{3}{4} \times 60 = 45$ minutes.

(C) The rate of flow of water through a tap is proportional to the area of its cross-section.
Since the tap is circular,the area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
Therefore,the rate of flow is proportional to the square of the diameter $(Rate \propto d^2)$.
Let $R_1$ be the rate of the first tap and $R_2$ be the rate of the second tap.
$R_1 \propto d^2$ and $R_2 \propto (2d)^2 = 4d^2$.
This means $R_2 = 4R_1$,which implies the second tap is $4$ times faster than the first tap.
Time taken is inversely proportional to the rate of flow $(Time \propto 1/Rate)$.
If the first tap takes $T_1 = 40$ $min$,the second tap will take $T_2 = T_1 / 4$.
$T_2 = 40 / 4 = 10$ $min$.

(A) Let the faster pipe fill the tank in $x$ $h$. Then,the slower pipe will fill it in $(x + 10)$ $h$.
The combined rate of the two pipes is $\frac{1}{x} + \frac{1}{x + 10} = \frac{1}{12}$.
Multiplying by $12x(x + 10)$,we get: $12(x + 10) + 12x = x(x + 10)$.
$12x + 120 + 12x = x^2 + 10x$.
$24x + 120 = x^2 + 10x$.
Rearranging into a quadratic equation: $x^2 - 14x - 120 = 0$.
Factoring the equation: $x^2 - 20x + 6x - 120 = 0$.
$x(x - 20) + 6(x - 20) = 0$.
$(x - 20)(x + 6) = 0$.
Since time cannot be negative,$x = 20$.
Therefore,the faster pipe takes $20$ $h$ to fill the tank.

(B) Pipe $A$ fills the cistern in $15 \text{ min}$,so its rate is $\frac{1}{15}$ of the tank per minute.
Pipe $B$ fills the cistern in $20 \text{ min}$,so its rate is $\frac{1}{20}$ of the tank per minute.
Both pipes are opened for $2 \text{ min}$. The work done by both pipes in $2 \text{ min}$ is $2 \times (\frac{1}{15} + \frac{1}{20}) = 2 \times (\frac{4+3}{60}) = 2 \times \frac{7}{60} = \frac{7}{30}$.
Remaining work to be filled by pipe $B$ is $1 - \frac{7}{30} = \frac{23}{30}$.
Time taken by pipe $B$ to fill the remaining part is $\frac{23/30}{1/20} = \frac{23}{30} \times 20 = \frac{46}{3} \text{ min}$.
Total time taken = (Time both pipes were open) + (Time pipe $B$ was open alone) = $2 + \frac{46}{3} = \frac{6+46}{3} = \frac{52}{3} \text{ min}$.

(D) Let the total capacity of the tank be the least common multiple of $24$ and $32$,which is $96$ units.
Efficiency of pipe $A = 96 / 24 = 4$ units/min.
Efficiency of pipe $B = 96 / 32 = 3$ units/min.
Let pipe $B$ be closed after $x$ minutes. Pipe $A$ remains open for the entire $9$ minutes.
Work done by pipe $A$ in $9$ minutes $= 9 \times 4 = 36$ units.
Remaining work to be done by pipe $B = 96 - 36 = 60$ units.
Time taken by pipe $B$ to complete $60$ units $= 60 / 3 = 20$ minutes.
Therefore,pipe $B$ should be closed after $20$ minutes.

(C) The work done by pipes $A, B$,and $C$ in a $3 \text{ min}$ cycle is:
$\frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{3 + 2 - 4}{60} = \frac{1}{60}$ part of the tank.
To avoid emptying the tank at the end,we calculate the time taken to reach near the capacity. Let $n$ be the number of cycles.
In $3n$ minutes,the tank filled is $\frac{n}{60}$.
We need to ensure that after $A$ and $B$ work,the tank is full. The capacity to be filled before the last two steps is $1 - (\frac{1}{20} + \frac{1}{30}) = 1 - \frac{5}{60} = 1 - \frac{1}{12} = \frac{11}{12}$.
$\frac{n}{60} \approx \frac{11}{12} \implies n = 55$.
In $55 \times 3 = 165 \text{ min}$,the tank filled is $\frac{55}{60} = \frac{11}{12}$ part.
Remaining part $= 1 - \frac{11}{12} = \frac{1}{12}$.
In the $166^{\text{th}} \text{ min}$,pipe $A$ fills $\frac{1}{20}$ part. Remaining $= \frac{1}{12} - \frac{1}{20} = \frac{5-3}{60} = \frac{2}{60} = \frac{1}{30}$.
In the $167^{\text{th}} \text{ min}$,pipe $B$ fills the remaining $\frac{1}{30}$ part.
Total time $= 165 + 1 + 1 = 167 \text{ min}$.

(D) Let the capacity of the tank be $1$ unit.
Normal rate of tap $A = \frac{1}{25}$ tank per minute.
Normal rate of tap $B = \frac{1}{20}$ tank per minute.
Due to improper opening,the actual rate of tap $A = \frac{1}{25} \times \frac{5}{6} = \frac{1}{30}$ tank per minute.
Actual rate of tap $B = \frac{1}{20} \times \frac{2}{3} = \frac{1}{30}$ tank per minute.
Combined rate of both taps $= \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$ tank per minute.
Therefore,the time taken to fill the tank $= \frac{1}{1/15} = 15$ $min$.

(B) The part of the tank filled by tap $A$ in $1$ min is $\frac{1}{20}$.
The part of the tank filled by tap $B$ in $1$ min is $\frac{1}{15}$.
The part of the tank filled by tap $C$ in $1$ min is $\frac{1}{12}$.
When all three taps are opened simultaneously,the part of the tank filled in $1$ min is $\frac{1}{20} + \frac{1}{15} + \frac{1}{12} = \frac{3 + 4 + 5}{60} = \frac{12}{60} = \frac{1}{5}$.
This means the total time required to fill the entire tank is $5$ min.
To fill $40 \%$ of the tank,the required time is $5 \times \frac{40}{100} = 5 \times 0.4 = 2$ min.

(D) Let the capacity of the cistern be the $LCM$ of $4, 5,$ and $2,$ which is $20$ units.
Efficiency of pipe $A = 20/4 = 5$ units/$min$.
Efficiency of pipe $B = 20/5 = 4$ units/$min$.
Efficiency of pipe $C$ (emptying) $= 20/2 = 10$ units/$min$.
In the first $2$ $min,$ pipes $A$ and $B$ are open. Total work done $= (5 + 4) \times 2 = 18$ units.
When the third pipe is opened,the net efficiency $= 5 + 4 - 10 = -1$ unit/$min$.
This means the cistern is being emptied at a rate of $1$ unit/$min$.
Time taken to empty the $18$ units $= 18 / 1 = 18$ $min$.

(C) The net rate of water flow into the tank when all taps are open is calculated by adding the rates of the inlet taps and subtracting the rate of the outlet tap.
Net rate $= 42 \ L/h + 56 \ L/h - 48 \ L/h = 50 \ L/h$.
Since the tank is filled in $16 \ h$ at this net rate,the total capacity of the tank is the product of the net rate and the time taken.
Capacity $= 50 \ L/h \times 16 \ h = 800 \ L$.

(C) The rate at which the boy pours water is $\frac{4}{3}$ litres per minute.
The rate at which the girl pours water is $\frac{3}{4}$ litres per minute.
Combined rate of the boy and the girl = $\frac{4}{3} + \frac{3}{4}$ litres per minute.
To add these fractions,find the common denominator,which is $12$: $\frac{16}{12} + \frac{9}{12} = \frac{25}{12}$ litres per minute.
Total volume to be filled = $100$ litres.
Time required = $\frac{\text{Total Volume}}{\text{Combined Rate}} = \frac{100}{\frac{25}{12}} = 100 \times \frac{12}{25} = 4 \times 12 = 48$ minutes.

(A) Let the inlet pipe fill the tank in $x$ hours.
The leakage is equivalent to half of the inflow,so the effective rate of filling becomes $1 - 1/2 = 1/2$ of the original rate.
Time taken to fill the tank with leakage = $2x$ hours.
According to the problem,the tank takes $36$ hours extra due to the leakage:
$2x - x = 36$
$x = 36$ hours.
Therefore,the inlet pipe alone can fill the tank in $36$ hours.

(D) The volume of water removed is $11 \ \text{litres}$. Since $1 \ \text{litre} = 1000 \ \text{cm}^3$,the volume removed is $11000 \ \text{cm}^3$.
The formula for the volume of a cylinder is $V = \pi r^2 h$,where $r$ is the radius and $h$ is the drop in water level.
Given the diameter is $25 \ \text{cm}$,the radius $r = 25/2 \ \text{cm}$.
Substituting the values: $11000 = (22/7) \times (25/2)^2 \times h$.
$11000 = (22/7) \times (625/4) \times h$.
Solving for $h$: $h = (11000 \times 7 \times 4) / (22 \times 625)$.
$h = (11000 \times 28) / 13750$.
$h = 308000 / 13750 = 22.4 \ \text{cm}$.
$22.4 \ \text{cm}$ can be written as $22 \frac{2}{5} \ \text{cm}$.

(C) The part of the tank filled by pipe $A$ in $1$ $hour$ is $\frac{1}{36}$.
The part of the tank filled by pipe $B$ in $1$ $hour$ is $\frac{1}{45}$.
When both pipes are opened simultaneously,the part of the tank filled in $1$ $hour$ is $\frac{1}{36} + \frac{1}{45}$.
Taking the least common multiple $(LCM)$ of $36$ and $45$,which is $180$,we get:
$\frac{5 + 4}{180} = \frac{9}{180} = \frac{1}{20}$.
Therefore,the total time required to fill the tank is the reciprocal of the work done in $1$ $hour$,which is $20$ $hours$.

(C) Let the diameters of the three pipes be $d_1 = 1 \text{ cm}$,$d_2 = 4/3 \text{ cm}$,and $d_3 = 2 \text{ cm}$.
Since the amount of water flowing is proportional to the square of the diameter,the rates of flow $R$ are $R_1 \propto 1^2$,$R_2 \propto (4/3)^2 = 16/9$,and $R_3 \propto 2^2 = 4$.
Given that the largest pipe $(d_3 = 2 \text{ cm})$ fills the cistern in $61 \text{ minutes}$,its rate is $R_3 = 1/61$ of the cistern per minute.
Since $R_3 = k \cdot 4 = 1/61$,the constant $k = 1/(61 \times 4)$.
Thus,$R_1 = 1 \times k = 1/(61 \times 4) = 1/244$ of the cistern per minute.
$R_2 = (16/9) \times k = (16/9) \times (1/244) = 4/(9 \times 61) = 4/549$ of the cistern per minute.
Total rate when running together: $R_{total} = R_1 + R_2 + R_3 = \frac{1}{244} + \frac{4}{549} + \frac{1}{61}$.
Converting to a common denominator ($244 = 4 \times 61$ and $549 = 9 \times 61$): $R_{total} = \frac{9 + 16 + 36}{36 \times 61} = \frac{61}{36 \times 61} = \frac{1}{36}$.
Therefore,the cistern will be filled in $36 \text{ minutes}$.

(C) The rate of filling by both pipes in $1$ $hr$ is $\frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$ of the tank.
Time taken to fill $\frac{1}{3}$ of the tank is $\frac{1/3}{1/12} = 4$ $hrs$.
After the tank is $\frac{1}{3}$ full,a leak develops. The leak removes $\frac{1}{3}$ of the water supplied by both pipes per hour.
Net rate of filling after the leak = $(\text{Rate of both pipes}) - (\text{Leakage rate}) = \frac{1}{12} - (\frac{1}{3} \times \frac{1}{12}) = \frac{1}{12} - \frac{1}{36} = \frac{3-1}{36} = \frac{2}{36} = \frac{1}{18}$ of the tank per hour.
The remaining part of the tank to be filled is $1 - \frac{1}{3} = \frac{2}{3}$.
Time taken to fill the remaining $\frac{2}{3}$ part = $\frac{2/3}{1/18} = \frac{2}{3} \times 18 = 12$ $hrs$.
Total time taken = $4$ $hrs + 12$ $hrs = 16$ $hrs$.

(B) The part of the tank filled by pipe $A$ in $1 \text{ hour}$ is $\frac{1}{4}$.
The part of the tank filled by pipe $B$ in $1 \text{ hour}$ is $\frac{1}{6}$.
The part of the tank emptied by pipe $C$ in $1 \text{ hour}$ is $\frac{1}{8}$.
When all three pipes are opened simultaneously,the net part of the tank filled in $1 \text{ hour}$ is $\frac{1}{4} + \frac{1}{6} - \frac{1}{8}$.
Taking the least common multiple $(LCM)$ of $4, 6,$ and $8$,which is $24$:
Net part filled in $1 \text{ hour} = \frac{6 + 4 - 3}{24} = \frac{7}{24}$.
Therefore,the total time required to fill the cistern is the reciprocal of the net part filled per hour:
$\text{Time} = \frac{24}{7} = 3 \frac{3}{7} \text{ hours}$.

(B) Let the faster pipe fill the reservoir in $x$ hours.
Then,the slower pipe will fill it in $(x + 5)$ hours.
Given that when both pipes work together,the reservoir is filled in $6$ hours.
Therefore,the rate of the faster pipe is $1/x$ and the rate of the slower pipe is $1/(x + 5)$.
The combined rate is $1/x + 1/(x + 5) = 1/6$.
Solving the equation: $(x + 5 + x) / (x(x + 5)) = 1/6$.
$6(2x + 5) = x^2 + 5x$.
$12x + 30 = x^2 + 5x$.
$x^2 - 7x - 30 = 0$.
$(x - 10)(x + 3) = 0$.
Since time cannot be negative,$x = 10$.
Thus,the faster pipe takes $10$ hours to fill the reservoir.

(B) Let the total capacity of the cistern be $1$ unit.
Combined rate of pipes $A, B$ and $C$ is $\frac{1}{6}$ units per hour.
In $2$ hours,the part filled by $A, B$ and $C$ together is $2 \times \frac{1}{6} = \frac{1}{3}$ units.
Remaining part to be filled $= 1 - \frac{1}{3} = \frac{2}{3}$ units.
This remaining $\frac{2}{3}$ part is filled by $A$ and $B$ in $8$ hours.
Therefore,the rate of $(A + B) = \frac{2/3}{8} = \frac{2}{24} = \frac{1}{12}$ units per hour.
Since the rate of $(A + B + C) = \frac{1}{6}$ and $(A + B) = \frac{1}{12}$,the rate of pipe $C$ is $\frac{1}{6} - \frac{1}{12} = \frac{2-1}{12} = \frac{1}{12}$ units per hour.
Thus,pipe $C$ alone can fill the cistern in $12$ hours.