Pipes and Cistern Questions in English

(C) Let the faster pipe fill the tank in $x$ minutes.
Then,the slower pipe fills the tank in $3x$ minutes.
According to the problem,the combined rate of the two pipes is $\frac{1}{x} + \frac{1}{3x} = \frac{1}{36}$.
Multiplying by $3x$,we get $3 + 1 = \frac{3x}{36}$.
$4 = \frac{x}{12}$,which implies $x = 48$ minutes.
The slower pipe takes $3x$ minutes to fill the tank.
Therefore,the time taken by the slower pipe is $3 \times 48 = 144$ minutes.

(B) The rate of filling for pipes $A, B$ and $C$ is $\frac{1}{30}, \frac{1}{20}$ and $\frac{1}{10}$ of the tank per minute respectively.
The total part of the tank filled in one minute by all three pipes is $\frac{1}{30} + \frac{1}{20} + \frac{1}{10} = \frac{2 + 3 + 6}{60} = \frac{11}{60}$.
The ratio of the volumes of solutions $P, Q$ and $R$ discharged into the tank is equal to the ratio of the rates of the pipes: $\frac{1}{30} : \frac{1}{20} : \frac{1}{10} = 2 : 3 : 6$.
Since the pipes are opened simultaneously and run for the same duration ($3$ minutes),the proportion of each solution in the mixture remains constant regardless of the total volume filled.
Therefore,the proportion of solution $R$ in the liquid is $\frac{6}{2 + 3 + 6} = \frac{6}{11}$.

(B) Let the total time taken to fill the cistern be $30$ minutes.
Let pipe $B$ be turned off after $x$ minutes.
Pipe $A$ works for the entire $30$ minutes,while pipe $B$ works for $x$ minutes.
The rate of pipe $A$ is $\frac{1}{37.5} = \frac{2}{75}$ cistern per minute.
The rate of pipe $B$ is $\frac{1}{45}$ cistern per minute.
According to the problem:
$\left(\frac{2}{75} \times 30\right) + \left(\frac{1}{45} \times x\right) = 1$
$\Rightarrow \frac{60}{75} + \frac{x}{45} = 1$
$\Rightarrow \frac{4}{5} + \frac{x}{45} = 1$
$\Rightarrow \frac{x}{45} = 1 - \frac{4}{5}$
$\Rightarrow \frac{x}{45} = \frac{1}{5}$
$\Rightarrow x = \frac{45}{5} = 9$ minutes.
Thus,pipe $B$ should be turned off after $9$ minutes.

(D) The efficiency of tap $A = \frac{1}{12}$ tank/hour,tap $B = \frac{1}{15}$ tank/hour,and tap $C = \frac{1}{20}$ tank/hour.
In the first hour,$A$ and $B$ are open. Part of the tank filled $= \frac{1}{12} + \frac{1}{15} = \frac{5+4}{60} = \frac{9}{60} = \frac{3}{20}$ tank.
In the second hour,$A$ and $C$ are open. Part of the tank filled $= \frac{1}{12} + \frac{1}{20} = \frac{5+3}{60} = \frac{8}{60} = \frac{2}{15}$ tank.
Thus,in a cycle of $2$ hours,the total part of the tank filled $= \frac{3}{20} + \frac{2}{15} = \frac{9+8}{60} = \frac{17}{60}$ tank.
In $3$ such cycles ($6$ hours),the part of the tank filled $= 3 \times \frac{17}{60} = \frac{17}{20}$ tank.
Remaining part to be filled $= 1 - \frac{17}{20} = \frac{3}{20}$ tank.
In the $7^{th}$ hour,$A$ and $B$ are opened. They fill $\frac{3}{20}$ tank in exactly $1$ hour.
Therefore,the total time taken $= 6 + 1 = 7$ hours.

(B) Let the time taken by pipe $C$ alone to empty the full cistern be $x$ $minutes$.
The rate of filling by pipe $A$ is $\frac{1}{60}$ of the cistern per $minute$.
The rate of filling by pipe $B$ is $\frac{1}{75}$ of the cistern per $minute$.
The rate of emptying by pipe $C$ is $\frac{1}{x}$ of the cistern per $minute$.
When all three pipes are opened simultaneously,the net rate of filling is $\frac{1}{50}$ of the cistern per $minute$.
Therefore,the equation is: $\frac{1}{60} + \frac{1}{75} - \frac{1}{x} = \frac{1}{50}$.
Rearranging the terms to solve for $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{60} + \frac{1}{75} - \frac{1}{50}$.
The least common multiple $(LCM)$ of $60, 75,$ and $50$ is $300$.
$\frac{1}{x} = \frac{5 + 4 - 6}{300} = \frac{3}{300} = \frac{1}{100}$.
Thus,$x = 100$ $minutes$.

(C) Let the efficiency of pipe $A$ be $1$ unit/hour.
Since pipe $B$ is twice as fast as $A$,the efficiency of $B$ is $2$ units/hour.
Since pipe $C$ is twice as fast as $B$,the efficiency of $C$ is $4$ units/hour.
Total efficiency of pipes $(A + B + C) = 1 + 2 + 4 = 7$ units/hour.
The tank is filled in $5$ hours,so the total capacity of the tank $= 7 \times 5 = 35$ units.
Time taken by pipe $A$ alone to fill the tank $= \frac{\text{Total Capacity}}{\text{Efficiency of } A} = \frac{35}{1} = 35$ hours.

(C) Let the time taken by $A$ alone to fill the cistern be $x$ hours.
Then,the time taken by $B$ alone to fill the cistern is $(x+6)$ hours.
According to the problem,the combined rate of filling is $\frac{1}{x} + \frac{1}{x+6} = \frac{1}{4}$.
Multiplying by $4x(x+6)$,we get $4(x+6) + 4x = x(x+6)$.
$4x + 24 + 4x = x^2 + 6x$.
$8x + 24 = x^2 + 6x$.
$x^2 - 2x - 24 = 0$.
Factoring the quadratic equation: $(x-6)(x+4) = 0$.
Since time cannot be negative,$x = 6$.
Therefore,pipe $A$ takes $6$ hours to fill the cistern alone.

(D) Let the capacity of the tank be $1$ unit.
The rate of the pump is $\frac{1}{2}$ tank per hour.
The effective rate with the leak is $\frac{1}{2 \frac{1}{3}} = \frac{1}{7/3} = \frac{3}{7}$ tank per hour.
Let the rate of the leak be $\frac{1}{x}$ tank per hour,where $x$ is the time taken to drain the tank.
The effective rate is given by: $\text{Rate of pump} - \text{Rate of leak} = \text{Effective rate}$.
$\frac{1}{2} - \frac{1}{x} = \frac{3}{7}$.
Rearranging the equation: $\frac{1}{x} = \frac{1}{2} - \frac{3}{7}$.
$\frac{1}{x} = \frac{7 - 6}{14} = \frac{1}{14}$.
Therefore,$x = 14$ hours.

(A) Part of the tank filled by pipe $A$ in $1$ minute $= \frac{1}{10}$.
Part of the tank emptied by pipe $B$ in $1$ minute $= \frac{1}{6}$.
Since pipe $B$ empties faster than pipe $A$ fills,the net effect when both are open is emptying the tank.
Net part of the tank emptied in $1$ minute $= \frac{1}{6} - \frac{1}{10} = \frac{5 - 3}{30} = \frac{2}{30} = \frac{1}{15}$.
This means the tank is emptied at a rate of $\frac{1}{15}$ of the total capacity per minute.
Since the tank is currently $\frac{2}{5}$ full,the time required to empty this $\frac{2}{5}$ portion is:
Time $= \frac{\text{Volume to be emptied}}{\text{Rate of emptying}} = \frac{2/5}{1/15} = \frac{2}{5} \times 15 = 6$ minutes.
Therefore,it will take $6$ minutes to empty the tank.

(C) Pipe $A$ can fill $\frac{1}{5}$ part of the tank in $1$ hour.
Pipe $B$ can fill $\frac{1}{6}$ part of the tank in $1$ hour.
Pipe $C$ can empty $\frac{1}{12}$ part of the tank in $1$ hour.
When all three pipes $A, B,$ and $C$ are opened together,the net part of the tank filled in $1$ hour is:
$= \frac{1}{5} + \frac{1}{6} - \frac{1}{12}$
Taking the least common multiple $(LCM)$ of $5, 6,$ and $12$,which is $60$:
$= \frac{12 + 10 - 5}{60} = \frac{17}{60}$ part per hour.
Therefore,the total time taken to fill the tank is the reciprocal of the work done in $1$ hour:
$= \frac{1}{17/60} = \frac{60}{17} \text{ hours} = 3 \frac{9}{17} \text{ hours}$.

(A) Pipe $A$ fills the tank in $20$ $minutes$,so its efficiency is $\frac{1}{20}$ of the tank per $minute$.
Pipe $B$ fills the tank in $30$ $minutes$,so its efficiency is $\frac{1}{30}$ of the tank per $minute$.
When both pipes are opened together,their combined efficiency is $\frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$ of the tank per $minute$.
Therefore,the time taken to fill the tank together is the reciprocal of the combined efficiency,which is $12$ $minutes$.

(C) The rate of filling for pipe $A$ is $\frac{1}{5}$ tank per hour.
The rate of filling for pipe $B$ is $\frac{1}{10}$ tank per hour.
The rate of filling for pipe $C$ is $\frac{1}{30}$ tank per hour.
When all three pipes are opened together,the combined rate of filling is:
$\text{Combined Rate} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30}$
To add these fractions,find the least common multiple $(LCM)$ of $5, 10,$ and $30$,which is $30$.
$\text{Combined Rate} = \frac{6}{30} + \frac{3}{30} + \frac{1}{30} = \frac{10}{30} = \frac{1}{3}$ tank per hour.
Therefore,the time taken to fill the tank is the reciprocal of the combined rate:
$\text{Time} = \frac{1}{1/3} = 3$ $hours$.

(D) Let the total capacity of the cistern be $1$ unit.
The rate of the filling tap is $\frac{1}{4}$ units per hour.
The rate of the emptying tap is $\frac{1}{9}$ units per hour.
When both taps are opened simultaneously,the net rate of filling is $\left(\frac{1}{4} - \frac{1}{9}\right)$ units per hour.
Net rate $= \frac{9 - 4}{36} = \frac{5}{36}$ units per hour.
The time taken to fill the cistern is the reciprocal of the net rate.
Time $= \frac{1}{\frac{5}{36}} = \frac{36}{5} = 7.2$ $hours$.

(B) Part of the tank filled by $A$ and $B$ in $1$ minute $= (1/18 + 1/9) = (1+2)/18 = 3/18 = 1/6$.
Part of the tank filled by $A$ and $B$ in $4$ minutes $= 4 \times (1/6) = 2/3$.
When $A, B,$ and $C$ are all open,the net rate of emptying the tank is $= (1/18 + 1/9) - 1/5 = 1/6 - 1/5 = (5-6)/30 = -1/30$.
The negative sign indicates that the tank is being emptied at a rate of $1/30$ of the tank per minute.
Time taken to empty the $2/3$ filled tank $= (2/3) / (1/30) = (2/3) \times 30 = 20$ minutes.

(C) Capacity of the tank = $10 \times 12 = 120$ litres.
To find the number of buckets needed when the capacity of each bucket is $8$ litres,we divide the total capacity of the tank by the capacity of the new bucket.
Number of buckets = $\frac{120}{8} = 15$.
Therefore,$15$ buckets are required.

(A) Let the rate of the first tap be $R_1 = \frac{1}{2}$ cistern per hour.
Let the rate of the second tap be $R_2 = -\frac{1}{3}$ cistern per hour (since it empties the cistern).
When both taps are opened,the net rate of filling is $R_{net} = R_1 + R_2 = \frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}$ cistern per hour.
Therefore,the time taken to fill the cistern is the reciprocal of the net rate,which is $6$ $hours$.

(A) Let the time taken by the first tap to fill the tank be $X = 25$ $minutes$ and the time taken by the second tap to empty the tank be $Y = 50$ $minutes$.
The part of the tank filled in $1$ $minute$ when both taps are open is given by:
$\frac{1}{X} - \frac{1}{Y} = \frac{1}{25} - \frac{1}{50} = \frac{2 - 1}{50} = \frac{1}{50}$.
Since the result is positive,the tank will be filled.
The total time taken to fill the tank is the reciprocal of the net rate:
$\text{Total time} = \frac{1}{1/50} = 50$ $minutes$.

(B) Let the capacity of the tank be $1$ unit.
Pipe $A$ fills the tank in $10$ minutes,so its rate is $\frac{1}{10}$ unit/minute.
Pipe $B$ empties the tank in $6$ minutes,so its rate is $-\frac{1}{6}$ unit/minute.
When both pipes are open,the net rate is $\frac{1}{10} - \frac{1}{6} = \frac{3-5}{30} = -\frac{2}{30} = -\frac{1}{15}$ unit/minute.
The negative sign indicates that the tank is being emptied.
The tank is currently $\frac{2}{5}$ full.
Time taken to empty the $\frac{2}{5}$ portion = $\frac{\text{Volume to be emptied}}{\text{Net rate}} = \frac{2/5}{1/15} = \frac{2}{5} \times 15 = 6$ minutes.
Therefore,it will take $6$ minutes to empty the tank.

(B) Let the time taken by tap $A$ be $X = 10$ $hours$ and by tap $B$ be $Y = 15$ $hours$.
The work done by tap $A$ in $1$ $hour$ is $1/10$.
The work done by tap $B$ in $1$ $hour$ is $1/15$.
When both taps are opened together,the work done in $1$ $hour$ is:
$(1/10 + 1/15) = (3+2)/30 = 5/30 = 1/6$.
Therefore,the time taken to fill the tank is the reciprocal of the work done in $1$ $hour$,which is $6$ $hours$.
Alternatively,using the formula $\frac{X \times Y}{X + Y}$:
$\frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6$ $hours$.

(C) Let the time taken by pipe $A$ be $X = 12 \text{ hours}$ and by pipe $B$ be $Y = 15 \text{ hours}$.
The part of the cistern emptied by pipe $A$ in $1 \text{ hour}$ is $\frac{1}{12}$.
The part of the cistern emptied by pipe $B$ in $1 \text{ hour}$ is $\frac{1}{15}$.
When both pipes are opened together,the part of the cistern emptied in $1 \text{ hour}$ is $\left(\frac{1}{12} + \frac{1}{15}\right) = \frac{5+4}{60} = \frac{9}{60} = \frac{3}{20}$.
Therefore,the total time taken to empty the cistern is the reciprocal of the work done in $1 \text{ hour}$,which is $\frac{20}{3} \text{ hours}$.
$\frac{20}{3} \text{ hours} = 6 \text{ hours } + \frac{2}{3} \times 60 \text{ minutes} = 6 \text{ hours } 40 \text{ minutes}$.

(A) Let the rates of the three pipes be $A, B,$ and $C$ respectively.
Rate of pipe $A = 1/10$ tank per hour.
Rate of pipe $B = 1/12$ tank per hour.
Rate of pipe $C$ (emptying) $= -1/20$ tank per hour.
When all three pipes are open simultaneously,the net rate of filling is:
$(1/10 + 1/12 - 1/20) \text{ tank per hour}$.
Taking the least common multiple $(LCM)$ of $10, 12,$ and $20$,which is $60$:
Net rate $= (6/60 + 5/60 - 3/60) = 8/60 = 2/15$ tank per hour.
Time taken to fill the tank $= 1 / (2/15) = 15/2 = 7.5$ hours.
$7.5$ hours is equal to $7$ hours and $30$ minutes.

(A) Let the time taken by pipes $A$,$B$,and $C$ be $X = 10$,$Y = 12$,and $Z = 15$ $hours$ respectively.
The work done by pipe $A$ in $1$ $hour = 1/10$.
The work done by pipe $B$ in $1$ $hour = 1/12$.
The work done by pipe $C$ in $1$ $hour = 1/15$.
Total work done by all three pipes in $1$ $hour = (1/10 + 1/12 + 1/15)$.
Taking the least common multiple $(LCM)$ of $10, 12, 15$,which is $60$:
Total work done in $1$ $hour = (6 + 5 + 4) / 60 = 15 / 60 = 1/4$.
Therefore,the total time taken to fill the cistern when all pipes are opened together is the reciprocal of the total work done in $1$ $hour$,which is $4$ $hours$.

(B) Let the capacity of the cistern be the Least Common Multiple $(LCM)$ of $24, 30,$ and $20$,which is $120$ units.
Efficiency of pipe $A = 120 / 24 = 5$ units/min.
Efficiency of pipe $B = 120 / 30 = 4$ units/min.
Efficiency of $(A + B - C) = 120 / 20 = 6$ units/min.
Here,$C$ is an outlet pipe,so its efficiency is subtracted.
$5 + 4 - \text{Efficiency of } C = 6$.
$9 - \text{Efficiency of } C = 6$.
Efficiency of $C = 9 - 6 = 3$ units/min.
Time taken by $C$ to empty the full cistern $= \text{Total capacity} / \text{Efficiency of } C = 120 / 3 = 40$ minutes.

(C) Let the rate of the inlet pipe be $1/8$ of the cistern per hour.
Due to the leak,the effective rate becomes $1/(8+2) = 1/10$ of the cistern per hour.
Let the rate of the leak be $1/L$ of the cistern per hour.
The net rate is given by: (Rate of inlet) - (Rate of leak) = (Effective rate).
$1/8 - 1/L = 1/10$.
$1/L = 1/8 - 1/10$.
$1/L = (5-4)/40 = 1/40$.
Therefore,the leak will empty the full cistern in $40$ $hours$.

(C) Let the capacity of the cistern be $C$ $litres$.
Rate of the leak $= \frac{C}{8}$ $litres/hour$.
Rate of the tap $= 6 \times 60 = 360$ $litres/hour$.
When both are open,the net rate of emptying is $\frac{C}{12}$ $litres/hour$.
The net rate is also given by (Rate of leak - Rate of tap).
So,$\frac{C}{8} - 360 = \frac{C}{12}$.
$\frac{C}{8} - \frac{C}{12} = 360$.
$\frac{3C - 2C}{24} = 360$.
$\frac{C}{24} = 360$.
$C = 360 \times 24 = 8640$ $litres$.

(D) Let the slower pipe take $x$ hours to fill the reservoir.
Then,the faster pipe takes $(x - 10)$ hours to fill the reservoir.
The rate of the slower pipe is $\frac{1}{x}$ reservoir per hour.
The rate of the faster pipe is $\frac{1}{x - 10}$ reservoir per hour.
When working together,the combined rate is $\frac{1}{12}$ reservoir per hour.
Therefore,$\frac{1}{x} + \frac{1}{x - 10} = \frac{1}{12}$.
Multiplying by $12x(x - 10)$,we get $12(x - 10) + 12x = x(x - 10)$.
$12x - 120 + 12x = x^2 - 10x$.
$24x - 120 = x^2 - 10x$.
$x^2 - 34x + 120 = 0$.
Factoring the quadratic equation: $(x - 30)(x - 4) = 0$.
This gives $x = 30$ or $x = 4$.
Since the faster pipe takes $(x - 10)$ hours,$x$ must be greater than $10$. Thus,$x = 30$.
The faster pipe takes $x - 10 = 30 - 10 = 20$ hours. Wait,re-evaluating: If $x=30$,faster pipe is $20$. If $x=4$,it is impossible. Let's re-check the calculation: $x^2 - 34x + 120 = 0$. Roots are $30$ and $4$. If $x=30$,faster pipe is $20$. If $x=20$,faster is $10$. Let's check $1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12$. Correct. The faster pipe takes $20$ hours. Since $20$ is not in the options,the answer is $\text{None of these}$.

(D) Let the time taken by pipe $B$ to fill the cistern be $x$ minutes.
Since pipe $A$ is $3$ times faster than pipe $B$,the time taken by pipe $A$ to fill the cistern is $\frac{x}{3}$ minutes.
According to the problem,pipe $A$ takes $32$ minutes less than pipe $B$:
$x - \frac{x}{3} = 32$
$\frac{2x}{3} = 32$
$x = 32 \times \frac{3}{2} = 48 \text{ minutes}$.
So,pipe $B$ takes $48$ minutes and pipe $A$ takes $48 / 3 = 16$ minutes.
Work done by pipe $A$ in $1$ minute = $\frac{1}{16}$.
Work done by pipe $B$ in $1$ minute = $\frac{1}{48}$.
Work done by both pipes in $1$ minute = $\frac{1}{16} + \frac{1}{48} = \frac{3+1}{48} = \frac{4}{48} = \frac{1}{12}$.
Therefore,both pipes together will fill the cistern in $12$ minutes.

(A) Pipe $A$ fills $\frac{1}{4}$ of the cistern in $1$ minute.
Pipe $B$ fills $\frac{1}{6}$ of the cistern in $1$ minute.
In a cycle of $2$ minutes (alternating $A$ then $B$),the part filled is $\frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$.
In $4$ minutes (two such cycles),the part filled is $2 \times \frac{5}{12} = \frac{10}{12} = \frac{5}{6}$.
The remaining part to be filled is $1 - \frac{5}{6} = \frac{1}{6}$.
In the $5^{th}$ minute,pipe $A$ operates. Since pipe $A$ fills $\frac{1}{4}$ in $60$ seconds,it will fill $\frac{1}{6}$ in $\frac{1/6}{1/4} \times 60 = \frac{4}{6} \times 60 = 40$ seconds.
Total time = $4$ minutes + $40$ seconds = $4 \, \text{min} \, 40 \, \text{s}$.

(C) Let the capacity of the tank be $1$ unit.
The rate of the first tap is $\frac{1}{10}$ tank per minute.
The rate of the second tap is $\frac{1}{12}$ tank per minute.
Let the rate of the third tap (emptying) be $x$ tank per minute.
When all three taps are open,the net rate is $\frac{1}{15}$ tank per minute.
Therefore,$\frac{1}{10} + \frac{1}{12} - x = \frac{1}{15}$.
Calculating the sum: $\frac{6+5}{60} - x = \frac{1}{15} \implies \frac{11}{60} - x = \frac{4}{60}$.
Thus,$x = \frac{11}{60} - \frac{4}{60} = \frac{7}{60}$ tank per minute.
The time taken by the third tap to empty the full tank is $\frac{1}{x} = \frac{60}{7}$ minutes.
$\frac{60}{7} \text{ minutes} = 8 \frac{4}{7} \text{ minutes} \approx 8 \text{ minutes and } 34 \text{ seconds}$.

(A) Work done by pipe $A$ in $1$ minute $= \frac{1}{15}$.
Work done by pipe $B$ in $1$ minute $= \frac{1}{18}$.
Work done by pipes $(A+B)$ in $6$ minutes $= 6 \times (\frac{1}{15} + \frac{1}{18}) = 6 \times (\frac{6+5}{90}) = 6 \times \frac{11}{90} = \frac{11}{15}$.
So,$\frac{11}{15}$ of the cistern is filled in $6$ minutes.
Now,all three pipes $(A+B+C)$ are opened. The net work done in $1$ minute $= \frac{1}{15} + \frac{1}{18} - \frac{1}{6} = \frac{6+5-15}{90} = \frac{-4}{90} = -\frac{2}{45}$.
The negative sign indicates that the cistern is being emptied at a rate of $\frac{2}{45}$ per minute.
Time taken to empty $\frac{11}{15}$ of the cistern $= \frac{11/15}{2/45} = \frac{11}{15} \times \frac{45}{2} = \frac{33}{2} = 16 \frac{1}{2}$ minutes.

(A) Let $A$ alone fill the reservoir in $x$ hours.
Then,$B$ alone can fill the reservoir in $(x+5)$ hours.
The rate of filling by $A$ is $\frac{1}{x}$ reservoir per hour.
The rate of filling by $B$ is $\frac{1}{x+5}$ reservoir per hour.
Together,they fill the reservoir in $6$ hours,so their combined rate is $\frac{1}{6}$ reservoir per hour.
Equation: $\frac{1}{x} + \frac{1}{x+5} = \frac{1}{6}$.
Multiplying by $6x(x+5)$,we get: $6(x+5) + 6x = x(x+5)$.
$6x + 30 + 6x = x^2 + 5x$.
$12x + 30 = x^2 + 5x$.
$x^2 - 7x - 30 = 0$.
Factoring the quadratic equation: $(x - 10)(x + 3) = 0$.
Since time cannot be negative,$x = 10$ hours.
Thus,pipe $A$ alone fills the reservoir in $10$ hours.

(A) The rate of filling by tap $A$ is $\frac{1}{20}$ of the cistern per minute.
The rate of filling by tap $B$ is $\frac{1}{25}$ of the cistern per minute.
Combined rate of filling when both taps are open is $\frac{1}{20} + \frac{1}{25} = \frac{5+4}{100} = \frac{9}{100}$ of the cistern per minute.
In $5$ $minutes$,the part filled is $\frac{9}{100} \times 5 = \frac{9}{20}$.
The remaining part to be filled is $1 - \frac{9}{20} = \frac{11}{20}$.
Since tap $B$ is turned off,tap $A$ fills the remaining part.
Time taken by tap $A$ to fill $\frac{11}{20}$ part is $\frac{11}{20} \div \frac{1}{20} = \frac{11}{20} \times 20 = 11$ $minutes$.

(B) Pipe $A$ fills $\frac{1}{6}$ of the tank in $1$ hour,and pipe $B$ fills $\frac{1}{8}$ of the tank in $1$ hour.
Combined work of $A$ and $B$ in $1$ hour $= \frac{1}{6} + \frac{1}{8} = \frac{4+3}{24} = \frac{7}{24}$.
Work done by both pipes in $1 \frac{1}{2}$ hours (or $\frac{3}{2}$ hours) $= \frac{7}{24} \times \frac{3}{2} = \frac{7}{16}$.
Remaining part of the tank to be filled $= 1 - \frac{7}{16} = \frac{9}{16}$.
Since pipe $A$ is turned off,pipe $B$ fills the remaining $\frac{9}{16}$ part.
Time taken by pipe $B$ to fill $\frac{9}{16}$ part $= \frac{9/16}{1/8} = \frac{9}{16} \times 8 = \frac{9}{2} = 4.5$ hours.
Total time taken to fill the tank $= 1.5$ hours (initial) $+ 4.5$ hours (remaining) $= 6$ hours.

(B) Let the capacity of the cistern be the Least Common Multiple $(LCM)$ of $12, 15,$ and $20,$ which is $60$ units.
Efficiency of the first tap = $60 / 12 = 5$ units/min.
Efficiency of the second tap = $60 / 15 = 4$ units/min.
When all pipes are open, the net efficiency = $60 / 20 = 3$ units/min.
Let the efficiency of the waste pipe be $x$ units/min.
Then, $5 + 4 - x = 3$.
$9 - x = 3$, which gives $x = 6$ units/min.
Time taken by the waste pipe to empty the full cistern = $\text{Total capacity} / \text{Efficiency of waste pipe} = 60 / 6 = 10$ minutes.

(B) Let the capacity of the cistern be the Least Common Multiple $(LCM)$ of $10, 15,$ and $18,$ which is $90$ units.
Efficiency of the first tap = $90 / 10 = 9$ units/min.
Efficiency of the second tap = $90 / 15 = 6$ units/min.
Combined efficiency of both taps = $9 + 6 = 15$ units/min.
When the waste pipe is open,the net efficiency = $90 / 18 = 5$ units/min.
Efficiency of the waste pipe = (Combined efficiency of taps) - (Net efficiency) = $15 - 5 = 10$ units/min.
Time taken by the waste pipe to empty the full cistern = $90 / 10 = 9$ minutes.

(C) Efficiency of pipe $X = \frac{1}{20}$ units/hour.
Efficiency of pipe $Y = \frac{1}{35}$ units/hour.
Total capacity of the tank = $LCM(20, 35) = 140$ units.
Efficiency of $X = 7$ units/hour and efficiency of $Y = 4$ units/hour.
Since they work on alternate hours starting with $Y$:
Cycle of $2$ hours ($Y$ then $X$) = $4 + 7 = 11$ units.
In $12$ cycles ($24$ hours),the work done = $12 \times 11 = 132$ units.
Remaining work = $140 - 132 = 8$ units.
In the $25^{\text{th}}$ hour,pipe $Y$ works and fills $4$ units. Remaining work = $8 - 4 = 4$ units.
In the $26^{\text{th}}$ hour,pipe $X$ works. Time taken by $X$ to fill $4$ units = $\frac{4}{7}$ hours.
Total time = $25 + \frac{4}{7} = \frac{175 + 4}{7} = \frac{179}{7}$ hours.

(C) Let the capacity of the tank be $C$ $gallons$.
Rate of the first inlet pipe $= \frac{C}{10}$ $gallons/hour$.
Rate of the second inlet pipe $= \frac{C}{12}$ $gallons/hour$.
Rate of the outlet pipe $= -80$ $gallons/hour$.
When all three pipes work together,the net rate is $\frac{C}{20}$ $gallons/hour$.
Therefore,$\frac{C}{10} + \frac{C}{12} - 80 = \frac{C}{20}$.
Rearranging the terms: $\frac{C}{10} + \frac{C}{12} - \frac{C}{20} = 80$.
Taking the least common multiple $(LCM)$ of $10, 12, 20$,which is $60$:
$\frac{6C + 5C - 3C}{60} = 80$.
$\frac{8C}{60} = 80$.
$\frac{2C}{15} = 80$.
$C = 80 \times \frac{15}{2} = 600$.
Thus,the capacity of the tank is $600$ $gallons$.

(B) Let the total capacity of the tank be the $LCM$ of $10$ and $12$,which is $60$ units.
Efficiency of tap $A = 60 / 10 = 6$ units/hour.
Efficiency of tap $B = 60 / 12 = 5$ units/hour.
The total time from $10 \text{ am}$ to $4 \text{ pm}$ is $6$ hours.
Since tap $B$ remains open for the entire duration of $6$ hours,the work done by tap $B = 5 \times 6 = 30$ units.
The remaining work to be done by tap $A = 60 - 30 = 30$ units.
The time for which tap $A$ must be open = $30 / 6 = 5$ hours.
Since tap $A$ was opened at $10 \text{ am}$,it should be closed after $5$ hours,which is at $3 \text{ pm}$.

(A) Pipe $P$ fills the tank in $p$ hours,so its efficiency (work done per hour) is $\frac{1}{p}$.
Pipe $Q$ empties the tank in $q$ hours,so its efficiency is $-\frac{1}{q}$ (negative because it removes water).
When both pipes are opened simultaneously,the net work done per hour is $\frac{1}{p} - \frac{1}{q}$.
If the tank is filled in $r$ hours,the net efficiency is also $\frac{1}{r}$.
Therefore,$\frac{1}{r} = \frac{1}{p} - \frac{1}{q}$.

(C) Efficiency of pipe $A = 1/6$ tank per hour.
Efficiency of pipe $B = 1/8$ tank per hour.
Combined efficiency of pipe $A$ and $B = (1/6 + 1/8) = (4+3)/24 = 7/24$ tank per hour.
In $2$ hours,the work done by both pipes $= 2 \times (7/24) = 7/12$ tank.
Remaining tank to be filled $= 1 - 7/12 = 5/12$ tank.
Pipe $B$ fills $1/8$ tank in $1$ hour.
Time taken by pipe $B$ to fill $5/12$ tank $= (5/12) / (1/8) = (5/12) \times 8 = 40/12 = 10/3 = 3 \frac{1}{3}$ hours.

(A) Pipe $A$ fills the tank in $4$ hours,so its efficiency is $12/4 = 3$ units/hour.
Pipe $B$ fills the tank in $6$ hours,so its efficiency is $12/6 = 2$ units/hour.
Total capacity of the tank = $\text{LCM}(4, 6) = 12$ units.
In the first hour,pipe $A$ fills $3$ units.
In the second hour,pipe $B$ fills $2$ units.
Thus,in a cycle of $2$ hours,the total work done is $3 + 2 = 5$ units.
After $2$ cycles ($4$ hours),the total work done is $5 \times 2 = 10$ units.
Remaining work = $12 - 10 = 2$ units.
In the $5^{\text{th}}$ hour,pipe $A$ works. Since pipe $A$ fills $3$ units in $1$ hour,it will take $2/3$ hours to fill the remaining $2$ units.
Total time = $4 + 2/3 = 4 \frac{2}{3}$ hours.

(B) Let the total capacity of the tank be the $LCM$ of $30, 45,$ and $36$,which is $180$ units.
Efficiency of $A = 180 / 30 = 6$ units/min.
Efficiency of $B = 180 / 45 = 4$ units/min.
Efficiency of $C = 180 / 36 = -5$ units/min (since it empties the tank).
In the first $12$ minutes,$A$ and $B$ work together: $(6 + 4) \times 12 = 10 \times 12 = 120$ units filled.
Remaining capacity to be filled $= 180 - 120 = 60$ units.
After $12$ minutes,$C$ is opened,so all three pipes work together: Efficiency $= 6 + 4 - 5 = 5$ units/min.
Time taken to fill the remaining $60$ units $= 60 / 5 = 12$ minutes.
Total time taken $= 12$ (initial) $+ 12$ (remaining) $= 24$ minutes.

(D) $1$. Calculate the efficiency of each pipe by finding the Least Common Multiple $(LCM)$ of $30, 45,$ and $36$,which is $180$ units.
$2$. Efficiency of $A = 180 / 30 = 6$ units/min.
$3$. Efficiency of $B = 180 / 45 = 4$ units/min.
$4$. Efficiency of $C = -180 / 36 = -5$ units/min (negative because it empties).
$5$. Pipes $A$ and $B$ work together for $12$ minutes: $(6 + 4) \times 12 = 10 \times 12 = 120$ units.
$6$. Remaining work to be filled $= 180 - 120 = 60$ units.
$7$. When $C$ is opened,the combined efficiency becomes $A + B + C = 6 + 4 - 5 = 5$ units/min.
$8$. Time taken to fill the remaining $60$ units $= 60 / 5 = 12$ minutes.
$9$. Total time taken $= 12$ (initial) $+ 12$ (remaining) $= 24$ minutes.

(B) The work done (filling the tank) is constant. The number of taps and the time taken are inversely proportional.
Let $M_1 = 9$ taps and $T_1 = 20$ minutes.
Let $M_2$ be the number of taps required to fill the tank in $T_2 = 15$ minutes.
Using the inverse proportion formula: $M_1 \times T_1 = M_2 \times T_2$
$9 \times 20 = M_2 \times 15$
$180 = M_2 \times 15$
$M_2 = \frac{180}{15} = 12$
Therefore,$12$ taps are required to fill the tank in $15$ minutes.

(D) The rate of dripping is $1 \text{ drop/sec}$.
Number of seconds in $1$ day $= 24 \times 60 \times 60 = 86400 \text{ seconds}$.
Total drops in $1$ day $= 86400 \text{ drops}$.
Given that $600 \text{ drops} = 100 \text{ ml}$,so $1 \text{ drop} = \frac{100}{600} \text{ ml} = \frac{1}{6} \text{ ml}$.
Water wasted in $1$ day $= 86400 \times \frac{1}{6} \text{ ml} = 14400 \text{ ml}$.
Since $1000 \text{ ml} = 1 \text{ litre}$,$14400 \text{ ml} = 14.4 \text{ litres}$.
Water wasted in $300$ days $= 14.4 \times 300 = 4320 \text{ litres}$.

(B) Let the time taken by the first pipe be $x$ hours.
Then,the time taken by the second pipe is $(x - 5)$ hours.
The time taken by the third pipe is $(x - 5) - 4 = (x - 9)$ hours.
According to the problem,the first and second pipes together fill the pool in the same time as the third pipe alone:
$\frac{1}{x} + \frac{1}{x - 5} = \frac{1}{x - 9}$
$(x - 5 + x) / (x(x - 5)) = 1 / (x - 9)$
$(2x - 5)(x - 9) = x(x - 5)$
$2x^2 - 18x - 5x + 45 = x^2 - 5x$
$x^2 - 18x + 45 = 0$
$(x - 15)(x - 3) = 0$
Since $x$ must be greater than $9$ (as the third pipe takes $x-9$ hours),we take $x = 15$.
Time taken by the first pipe = $15$ hours,second pipe = $10$ hours,third pipe = $6$ hours.
Combined rate of second and third pipes = $\frac{1}{10} + \frac{1}{6} = \frac{3 + 5}{30} = \frac{8}{30} = \frac{4}{15}$ pools per hour.
Time taken = $\frac{15}{4} = 3.75$ hours.

(D) The pool is filled at a constant rate.
It takes $8$ hours to fill $\frac{3}{5}$ of the pool's capacity.
Remaining capacity to be filled = $1 - \frac{3}{5} = \frac{2}{5}$ of the capacity.
Let the total time to fill the pool be $T$ hours.
Since the rate is constant,the time taken is proportional to the fraction filled:
$\frac{3}{5} \text{ capacity} \rightarrow 8 \text{ hours}$
$\frac{2}{5} \text{ capacity} \rightarrow x \text{ hours}$
Using the ratio method: $x = \frac{8 \times (2/5)}{3/5} = \frac{8 \times 2}{3} = \frac{16}{3} \text{ hours}$.
$\frac{16}{3} \text{ hours} = 5 \frac{1}{3} \text{ hours}$.
Since $1 \text{ hour} = 60 \text{ minutes}$,$\frac{1}{3} \text{ hour} = \frac{1}{3} \times 60 = 20 \text{ minutes}$.
Therefore,the remaining time is $5 \text{ hours } 20 \text{ minutes}$.

(D) The rate of the pump filling the tank is $\frac{1}{2}$ of the tank per hour.
Due to the leak,the effective rate of filling becomes $\frac{1}{2 \frac{1}{3}} = \frac{1}{7/3} = \frac{3}{7}$ of the tank per hour.
The rate at which the leak drains the tank is the difference between the pump's rate and the effective rate:
$\text{Leak rate} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14}$ of the tank per hour.
Therefore,the leak can drain the entire tank in $14$ $hours$.

(A) The part of the cistern filled by pipe $A$ in $1$ $hour$ is $\frac{1}{3}$.
The part of the cistern filled by pipe $B$ in $1$ $hour$ is $\frac{1}{5}$.
The part of the cistern emptied by pipe $C$ in $1$ $hour$ is $\frac{1}{2}$.
When all three pipes are open,the net part of the cistern filled in $1$ $hour$ is:
$\frac{1}{3} + \frac{1}{5} - \frac{1}{2} = \frac{10 + 6 - 15}{30} = \frac{1}{30}$.
Since the net work done in $1$ $hour$ is $\frac{1}{30}$ of the cistern,the total time required to fill the cistern is $30$ $hours$.

(C) Let the capacity of the cistern be $C$ litres.
Rate of the first pipe $= \frac{C}{24}$ litres/minute.
Rate of the second pipe $= \frac{C}{40}$ litres/minute.
Rate of the waste pipe $= -30$ litres/minute.
When all three pipes are open,the cistern fills in $1$ hour ($60$ minutes).
Therefore,the net rate of filling is $\frac{C}{60}$ litres/minute.
Equation: $\frac{C}{24} + \frac{C}{40} - 30 = \frac{C}{60}$.
Rearranging the terms: $\frac{C}{24} + \frac{C}{40} - \frac{C}{60} = 30$.
Taking the $LCM$ of $24, 40,$ and $60$,which is $120$:
$\frac{5C + 3C - 2C}{120} = 30$.
$\frac{6C}{120} = 30$.
$\frac{C}{20} = 30$.
$C = 30 \times 20 = 600$ litres.
Thus,the capacity of the cistern is $600$ litres.