Pipes and Cistern Questions in English

(D) Let the pipe $B$ be closed after $x$ minutes.
Pipe $A$ remains open for the entire duration of $18$ minutes.
Work done by pipe $A$ in $18$ minutes = $\frac{18}{24} = \frac{3}{4}$.
Remaining work to be done by pipe $B$ = $1 - \frac{3}{4} = \frac{1}{4}$.
Pipe $B$ fills the tank in $32$ minutes,so the time taken by $B$ to do $\frac{1}{4}$ of the work is $x = 32 \times \frac{1}{4} = 8$ minutes.
Therefore,pipe $B$ should be closed after $8$ minutes.

(C) Let the capacity of the tank be $60$ units ($LCM$ of $10, 15, 12$).
Efficiency of the first tap $= 60 / 10 = 6$ units/$hr$.
Efficiency of the second tap $= 60 / 15 = 4$ units/$hr$.
Combined efficiency of all three taps $= 60 / 12 = 5$ units/$hr$.
Let the efficiency of the third tap be $x$ units/$hr$.
Since the third tap empties the tank,its efficiency is negative.
$6 + 4 - x = 5$
$10 - x = 5$
$x = 5$ units/$hr$.
Time taken by the third tap to empty the tank $= 60 / 5 = 12$ $hrs$.

(D) Let the work done by pipe $A$ in $1$ minute be $a$ and by pipe $B$ be $b$.
Given that $(a + b) = \frac{1}{36}$.
In $30$ minutes,the part of the tank filled by both pipes is $30(a + b) = 30 \times \frac{1}{36} = \frac{5}{6}$.
The remaining part of the tank is $1 - \frac{5}{6} = \frac{1}{6}$.
This remaining part is filled by pipe $A$ alone in $(40 - 30) = 10$ minutes.
So,$10a = \frac{1}{6}$,which means $a = \frac{1}{60}$.
Now,substitute $a$ in the first equation: $\frac{1}{60} + b = \frac{1}{36}$.
$b = \frac{1}{36} - \frac{1}{60} = \frac{5 - 3}{180} = \frac{2}{180} = \frac{1}{90}$.
Therefore,pipe $B$ alone can fill the tank in $90$ minutes.

(D) Let the total capacity of the tank be $180$ units ($LCM$ of $45$ and $60$).
Efficiency of the first pipe (filling) $= 180 / 45 = +4$ units/min.
Efficiency of the second pipe (emptying) $= 180 / 60 = -3$ units/min.
In a cycle of $2$ minutes (first pipe on for $1$ min,second pipe on for $1$ min),the net work done $= 4 - 3 = 1$ unit.
To avoid overflowing,we consider the capacity before the last filling step: $180 - 4 = 176$ units.
It takes $176$ cycles to fill $176$ units,which equals $176 \times 2 = 352$ minutes.
In the next minute ($353^{\text{rd}}$ minute),the first pipe fills the remaining $4$ units.
Total time $= 352 + 1 = 353$ minutes $= 5$ hours and $53$ minutes.

(D) Let the total capacity of the tank be $60$ units ($LCM$ of $20$ and $60$).
Efficiency of the $1$st tap $= 60 / 20 = 3$ units/$min$.
Efficiency of the $2$nd tap $= 60 / 60 = 1$ unit/$min$.
In $5$ $min$,both taps together fill $= (3 + 1) \times 5 = 20$ units.
Remaining capacity $= 60 - 20 = 40$ units.
Since the $1$st tap is shut off,the $2$nd tap fills the remaining $40$ units.
Time taken by the $2$nd tap $= 40 / 1 = 40$ $min$.

(B) Pipe $A$ fills $1/3$ of the cistern in $1 \, h$. Pipe $B$ fills $1/4$ of the cistern in $1 \, h$. Pipe $C$ empties $1$ full cistern in $1 \, h$.
At $4:00 \, pm$, pipe $A$ has worked for $1 \, h$, filling $1/3$ of the cistern.
At $5:00 \, pm$, pipe $A$ has worked for $2 \, h$ and pipe $B$ has worked for $1 \, h$. Total filled part $= (2 \times 1/3) + (1 \times 1/4) = 2/3 + 1/4 = 11/12$.
From $5:00 \, pm$ onwards, all three pipes $A, B,$ and $C$ are open. The net work done per hour $= 1/3 + 1/4 - 1 = (4+3-12)/12 = -5/12$. The negative sign indicates the cistern is being emptied at a rate of $5/12$ per hour.
Time taken to empty the $11/12$ part $= (11/12) / (5/12) = 11/5 \, h = 2.2 \, h$.
$2.2 \, h = 2 \, h$ and $0.2 \times 60 \, min = 2 \, h \, 12 \, min$.
Starting from $5:00 \, pm$, adding $2 \, h \, 12 \, min$ gives $7:12 \, pm$.

(C) The rate of filling for pipes $A, B,$ and $C$ are $\frac{1}{30}, \frac{1}{20},$ and $\frac{1}{10}$ of the tank per minute,respectively.
The total part of the tank filled in $1 \ min$ by all three pipes is $\frac{1}{30} + \frac{1}{20} + \frac{1}{10} = \frac{2 + 3 + 6}{60} = \frac{11}{60}$.
In $3 \ min$,the total part of the tank filled is $3 \times \frac{11}{60} = \frac{33}{60} = \frac{11}{20}$.
The amount of solution $R$ contributed by pipe $C$ in $3 \ min$ is $3 \times \frac{1}{10} = \frac{3}{10}$.
The fraction of solution $R$ in the total liquid in the tank after $3 \ min$ is $\frac{\text{Amount of } R}{\text{Total liquid}} = \frac{3/10}{11/20} = \frac{3}{10} \times \frac{20}{11} = \frac{6}{11}$.

(B) Let the total capacity of the cistern be $1$ unit.
Combined efficiency of pipes $A, B,$ and $C$ is $(A+B+C) = \frac{1}{6}$ units/hr.
In $2$ hours,the work done by $A, B,$ and $C$ is $2 \times \frac{1}{6} = \frac{1}{3}$ units.
Remaining work $= 1 - \frac{1}{3} = \frac{2}{3}$ units.
This remaining work is done by $A$ and $B$ in $8$ hours.
So,the efficiency of $(A+B) = \frac{2/3}{8} = \frac{2}{24} = \frac{1}{12}$ units/hr.
Now,the efficiency of pipe $C = (A+B+C) - (A+B) = \frac{1}{6} - \frac{1}{12} = \frac{2-1}{12} = \frac{1}{12}$ units/hr.
Therefore,the time taken by pipe $C$ to fill the cistern alone is $\frac{1}{1/12} = 12$ $hrs$.

(D) Let the work done by pipes $A, B$ and $C$ in $1$ $hr$ be $a, b$ and $c$ respectively.
Given:
$a + b = \frac{1}{6}$
$b + c = \frac{1}{10}$
$c + a = \frac{1}{7.5} = \frac{1}{15/2} = \frac{2}{15}$
Adding these three equations:
$2(a + b + c) = \frac{1}{6} + \frac{1}{10} + \frac{2}{15}$
$2(a + b + c) = \frac{5 + 3 + 4}{30} = \frac{12}{30} = \frac{2}{5}$
$a + b + c = \frac{1}{5}$
Thus,all three pipes together will fill the tank in $5$ $hrs$.

(A) Let the total time taken to fill the tank be $x$ hours.
Rate of pipe $A = 1/15$ tank/hour.
Rate of pipe $B = 1/20$ tank/hour.
Rate of pipe $C = -1/25$ tank/hour (emptying).
Since pipe $C$ is closed after $10$ hours,it works for $10$ hours,while pipes $A$ and $B$ work for the entire duration $x$.
The equation is: $x(1/15 + 1/20) - 10(1/25) = 1$.
Solving for $x$: $x((4+3)/60) - 10/25 = 1$.
$x(7/60) - 0.4 = 1$.
$x(7/60) = 1.4$.
$x = 1.4 \times (60/7) = 1.4 \times 8.57...$ (Wait,recalculating).
$x(7/60) = 1 + 0.4 = 1.4$.
$x = 1.4 \times 60 / 7 = 84 / 7 = 12$ hours.
Thus,the tank is filled in $12$ hours.

(D) Work done by $A$ in $1$ hour $= \frac{1}{10}$.
Work done by $A$ and $B$ in the second hour $= \frac{1}{10} + \frac{1}{12} = \frac{6+5}{60} = \frac{11}{60}$.
Total work done in the first $2$ hours $= \frac{1}{10} + \frac{11}{60} = \frac{6+11}{60} = \frac{17}{60}$.
Remaining work $= 1 - \frac{17}{60} = \frac{43}{60}$.
Work done by $A, B$ and $C$ in $1$ hour $= \frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac{6+5+4}{60} = \frac{15}{60} = \frac{1}{4}$.
Time taken to complete the remaining work $= \frac{43/60}{1/4} = \frac{43}{60} \times 4 = \frac{43}{15} = 2 \text{ hours } 52 \text{ minutes}$.
Total time $= 2 \text{ hours} + 2 \text{ hours } 52 \text{ minutes} = 4 \text{ hours } 52 \text{ minutes}$.

(A) The work done by pipes $A, B, C$ in a cycle of $3$ minutes is:
$\frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{3 + 2 - 4}{60} = \frac{1}{60}$ part of the tank.
To avoid overfilling,we consider the capacity filled before the last cycle. The tank capacity is $1$. Pipe $A$ fills $\frac{1}{20}$ and $B$ fills $\frac{1}{30}$.
We need to reach a point where the remaining part is $\le \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12}$.
In $165$ minutes ($55$ cycles of $3$ minutes),the part filled is $55 \times \frac{1}{60} = \frac{55}{60} = \frac{11}{12}$.
Remaining part $= 1 - \frac{11}{12} = \frac{1}{12}$.
In the $166$th minute,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 $167$th minute,pipe $B$ fills $\frac{1}{30}$ part. Thus,the tank is full.
Total time $= 165 + 1 + 1 = 167$ minutes.

(C) Let the efficiency of pipe $A$ be $1$ unit per hour.
Since $B$ is twice as fast as $A$,the efficiency of $B = 2$ units per hour.
Since $C$ is twice as fast as $B$,the efficiency of $C = 2 \times 2 = 4$ units per hour.
Total efficiency of pipes $(A + B + C) = 1 + 2 + 4 = 7$ units per 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.

(D) Pipe $A$ fills the tank in $15$ minutes,so its rate is $\frac{1}{15}$ tank per minute.
Pipe $B$ fills the tank in $20$ minutes,so its rate is $\frac{1}{20}$ tank per minute.
Both pipes are opened together for $4$ minutes. The work done by both in $4$ minutes is $4 \times (\frac{1}{15} + \frac{1}{20}) = 4 \times (\frac{4+3}{60}) = 4 \times \frac{7}{60} = \frac{7}{15}$.
The remaining part of the tank to be filled is $1 - \frac{7}{15} = \frac{8}{15}$.
Pipe $B$ fills this remaining part at a rate of $\frac{1}{20}$ per minute.
Time taken by pipe $B$ to fill the remaining part $= \frac{8/15}{1/20} = \frac{8}{15} \times 20 = \frac{32}{3} = 10 \text{ minutes and } 40 \text{ seconds}$.
Total time taken $= 4 \text{ minutes} + 10 \text{ minutes } 40 \text{ seconds} = 14 \text{ minutes } 40 \text{ seconds}$.

(D) Efficiency of $A = \frac{1}{12}$ tank/hour,$B = \frac{1}{15}$ tank/hour,$C = \frac{1}{20}$ tank/hour.
In the first hour,$A$ and $B$ are open: Work done $= \frac{1}{12} + \frac{1}{15} = \frac{5+4}{60} = \frac{9}{60}$ tank.
In the second hour,$A$ and $C$ are open: Work done $= \frac{1}{12} + \frac{1}{20} = \frac{5+3}{60} = \frac{8}{60}$ tank.
In a cycle of $2$ hours,the total work done $= \frac{9}{60} + \frac{8}{60} = \frac{17}{60}$ tank.
In $6$ hours ($3$ cycles),the work done $= 3 \times \frac{17}{60} = \frac{51}{60}$ tank.
Remaining work $= 1 - \frac{51}{60} = \frac{9}{60}$ tank.
In the $7$th hour,$A$ and $B$ are open. They fill $\frac{9}{60}$ tank in exactly $1$ hour.
Total time $= 6 + 1 = 7$ hours.

(A) Let the filling capacity of the pump be $x m^3/min$.
The emptying capacity of the pump is $(x + 10) m^3/min$.
The time taken to fill the tank is $T_f = \frac{2400}{x}$ minutes.
The time taken to empty the tank is $T_e = \frac{2400}{x+10}$ minutes.
According to the problem,$T_f - T_e = 8$.
$\frac{2400}{x} - \frac{2400}{x+10} = 8$
Divide by $8$: $\frac{300}{x} - \frac{300}{x+10} = 1$
$300(x + 10) - 300x = x(x + 10)$
$300x + 3000 - 300x = x^2 + 10x$
$x^2 + 10x - 3000 = 0$
$(x + 60)(x - 50) = 0$
Since capacity cannot be negative,$x = 50$.
Thus,the filling capacity of the pump is $50 m^3/min$.

(D) Let the total capacity of the tank be $V$ litres.
Rate of the leak $= \frac{V}{8}$ litres per hour.
Rate of the inlet pipe $= 6 \times 60 = 360$ litres per hour.
When the tank is full and both are active,the net rate of emptying is $\frac{V}{12}$ litres per hour.
Therefore,the net rate is given by: $\text{Leak rate} - \text{Inlet rate} = \text{Net emptying rate}$.
$\frac{V}{8} - 360 = \frac{V}{12}$.
$\frac{V}{8} - \frac{V}{12} = 360$.
$\frac{3V - 2V}{24} = 360$.
$\frac{V}{24} = 360$.
$V = 360 \times 24 = 8640$ litres.
Thus,the tank holds $8640$ litres.

(C) Let the capacity of the tank be $C$ gallons.
Rate of first pipe $= \frac{C}{20}$ gallons/min.
Rate of second pipe $= \frac{C}{24}$ gallons/min.
Rate of waste pipe $= -3$ gallons/min.
When all three pipes work together,the net rate is $\frac{C}{15}$ gallons/min.
So,$\frac{C}{20} + \frac{C}{24} - 3 = \frac{C}{15}$.
$\frac{C}{20} + \frac{C}{24} - \frac{C}{15} = 3$.
Taking $LCM$ of $20, 24, 15$ which is $120$:
$\frac{6C + 5C - 8C}{120} = 3$.
$\frac{3C}{120} = 3$.
$\frac{C}{40} = 3$.
$C = 120$ gallons.

(C) Let the total capacity of the tank be $1$ unit.
Combined work rate of $A, B$ and $C$ is $(A + B + C) = \frac{1}{6}$ unit/hour.
Work done by $A, B$ and $C$ in $2$ hours $= 2 \times \frac{1}{6} = \frac{1}{3}$ unit.
Remaining part of the tank $= 1 - \frac{1}{3} = \frac{2}{3}$ unit.
This remaining $\frac{2}{3}$ part is filled by $A$ and $B$ in $7$ hours.
Therefore,the work rate of $(A + B) = \frac{2/3}{7} = \frac{2}{21}$ unit/hour.
Now,the work rate of $C = (A + B + C) - (A + B) = \frac{1}{6} - \frac{2}{21}$.
Taking the $LCM$ of $6$ and $21$,which is $42$:
Work rate of $C = \frac{7 - 4}{42} = \frac{3}{42} = \frac{1}{14}$ unit/hour.
Thus,$C$ alone can fill the tank in $14$ hours.

(A) Let the pipe $B$ be closed after $x$ minutes.
Pipe $A$ remains open for the entire $20$ minutes.
Work done by pipe $A$ in $20$ minutes = $\frac{20}{24} = \frac{5}{6}$.
Remaining work to be done by pipe $B$ = $1 - \frac{5}{6} = \frac{1}{6}$.
Pipe $B$ fills the tank in $48$ minutes,so the time taken by $B$ to do $\frac{1}{6}$ of the work is $x = 48 \times \frac{1}{6} = 8$ minutes.
Therefore,pipe $B$ should be closed after $8$ minutes.

(B) Let the first tap be closed after $x$ minutes.
The rate of the first tap is $1/20$ tank per minute,and the rate of the second tap is $1/30$ tank per minute.
The first tap worked for $x$ minutes,and the second tap worked for the entire $18$ minutes.
According to the problem,the total work done is equal to filling one full tank:
$\frac{x}{20} + \frac{18}{30} = 1$
Simplify the equation:
$\frac{x}{20} + \frac{3}{5} = 1$
Subtract $3/5$ from both sides:
$\frac{x}{20} = 1 - \frac{3}{5}$
$\frac{x}{20} = \frac{2}{5}$
Solve for $x$:
$x = \frac{2}{5} \times 20$
$x = 8$ minutes.
Thus,the first tap was closed after $8$ minutes.

(D) Efficiency of tap $P = \frac{1}{12}$ tank per minute.
Efficiency of tap $Q = \frac{1}{15}$ tank per minute.
Combined efficiency of $P$ and $Q = \frac{1}{12} + \frac{1}{15} = \frac{5+4}{60} = \frac{9}{60} = \frac{3}{20}$ tank per minute.
Work done by both in $3$ minutes $= 3 \times \frac{3}{20} = \frac{9}{20}$ tank.
Remaining part of the tank $= 1 - \frac{9}{20} = \frac{11}{20}$ tank.
Time taken by $Q$ to fill the remaining $\frac{11}{20}$ part $= \frac{11/20}{1/15} = \frac{11}{20} \times 15 = \frac{11 \times 3}{4} = \frac{33}{4} = 8 \frac{1}{4}$ $min$.

(D) Let the faster pipe take $x$ $min$ to fill the tank. Then,the slower pipe takes $3x$ $min$ to fill it.
The rate of the faster pipe is $\frac{1}{x}$ tank per minute,and the rate of the slower pipe is $\frac{1}{3x}$ tank per minute.
Combined rate: $\frac{1}{x} + \frac{1}{3x} = \frac{1}{36}$.
Solving for $x$: $\frac{3+1}{3x} = \frac{1}{36} \Rightarrow \frac{4}{3x} = \frac{1}{36}$.
$3x = 4 \times 36 = 144$.
Thus,the slower pipe takes $144$ $min$ to fill the tank alone.
$144$ $min = 2$ $hours$ $24$ $min$.

(C) Let the total capacity of the tank be $C$ units.
Rate of the first tap (filling) $= \frac{C}{12}$ units per hour.
Rate of the second tap (emptying) $= \frac{C/2}{10} = \frac{C}{20}$ units per hour.
When both taps are opened,the net rate of filling $= \frac{C}{12} - \frac{C}{20} = \frac{5C - 3C}{60} = \frac{2C}{60} = \frac{C}{30}$ units per hour.
To fill half of the tank (i.e.,$C/2$ units),the time required $= \frac{C/2}{C/30} = \frac{C}{2} \times \frac{30}{C} = 15$ $hours$.

(A) Efficiency of filling pipe $= \frac{1}{40}$ drum/min.
Efficiency of outlet pipe $= -\frac{1}{60}$ drum/min.
Initial state of the drum $= \frac{2}{3}$ full.
Outlet pipe is opened for $15$ min,so the amount emptied $= 15 \times \frac{1}{60} = \frac{1}{4}$ part.
Remaining amount in the drum $= \frac{2}{3} - \frac{1}{4} = \frac{8-3}{12} = \frac{5}{12}$ part.
Amount to be filled to make the drum full $= 1 - \frac{5}{12} = \frac{7}{12}$ part.
Time required for the filling pipe to fill the remaining $\frac{7}{12}$ part $= \frac{7/12}{1/40} = \frac{7}{12} \times 40 = \frac{7 \times 10}{3} = \frac{70}{3} = 23 \frac{1}{3} \text{ min}$.

(NONE) Let the capacity of the tank be $V$ gallons.
The rate of tap $A$ is $\frac{V}{20}$ gallons/min.
The rate of tap $B$ is $\frac{V}{24}$ gallons/min.
The rate of tap $C$ is $3$ gallons/min (emptying).
When all three are opened,the net rate is $\frac{V}{15}$ gallons/min.
Therefore,$\frac{V}{20} + \frac{V}{24} - 3 = \frac{V}{15}$.
Rearranging the terms: $\frac{V}{20} + \frac{V}{24} - \frac{V}{15} = 3$.
Taking the $LCM$ of $20, 24,$ and $15$,which is $120$:
$\frac{6V + 5V - 8V}{120} = 3$.
$\frac{3V}{120} = 3$.
$\frac{V}{40} = 3$.
$V = 120 \times 3 / 3 = 120$ is incorrect based on the previous logic; let's re-calculate: $\frac{3V}{120} = 3 \implies V = 120$.

(D) Let the total capacity of the tank be the $LCM$ of $4, 8, 12,$ and $10$,which is $120$ units.
Efficiency of $P = 120/4 = 30$ units/hr.
Efficiency of $Q = 120/8 = 15$ units/hr.
Efficiency of $R = 120/12 = 10$ units/hr.
Efficiency of $S = -120/10 = -12$ units/hr (negative because it empties).
$(i)$ $Q$ opened alone: Time $= 120/15 = 8$ hours.
(ii) $P$ and $S$ opened: Net efficiency $= 30 - 12 = 18$ units/hr. Time $= 120/18 = 6.67$ hours.
(iii) $P, R$ and $S$ opened: Net efficiency $= 30 + 10 - 12 = 28$ units/hr. Time $= 120/28 = 4.28$ hours.
(iv) $P, Q$ and $S$ opened: Net efficiency $= 30 + 15 - 12 = 33$ units/hr. Time $= 120/33 = 3.64$ hours.
Comparing the times: $3.64 < 4.28 < 6.67 < 8$. Therefore,option $(d)$ takes the least time.

(D) The rate at which the first pipe fills the tank is $\frac{1}{x}$ tank per hour.
The rate at which the second pipe empties the tank is $\frac{1}{y}$ tank per hour.
When both pipes are open,the net rate of filling the tank is $\left( \frac{1}{x} - \frac{1}{y} \right)$ tank per hour.
Simplifying the expression: $\frac{1}{x} - \frac{1}{y} = \frac{y - x}{xy}$ tank per hour.
The time required to fill the tank is the reciprocal of the net rate.
Therefore,the required time is $\frac{xy}{y - x} \text{ h}$.

(B) Let the rate of the filling pipe be $R_f = \frac{1}{3}$ tank per hour.
Let the rate of the leak be $R_l$ tank per hour.
The effective rate when both are working is $R_{eff} = \frac{1}{3.5} = \frac{1}{7/2} = \frac{2}{7}$ tank per hour.
Since $R_{eff} = R_f - R_l$,we have $R_l = R_f - R_{eff}$.
$R_l = \frac{1}{3} - \frac{2}{7} = \frac{7 - 6}{21} = \frac{1}{21}$ tank per hour.
Therefore,the time taken by the leak to empty the full tank is $\frac{1}{1/21} = 21 \, h$.

(B) Let the capacity of the tank be $1$ unit.
Rate of the $1$st pipe (filling) $= 1/12$ tank per minute.
Rate of the $2$nd pipe (emptying) $= 1/20$ tank per minute.
When both pipes are open,the net rate of filling $= (1/12 - 1/20)$ tank per minute.
$= (5 - 3) / 60 = 2 / 60 = 1/30$ tank per minute.
To fill $1/2$ of the tank,the time required $= (1/2) / (1/30) = 30 / 2 = 15$ $min$.

(B) Let the total capacity of the cistern be the least common multiple of $3, 4,$ and $5$,which is $60$ units.
Efficiency of the first tap $= 60 / 3 = 20$ units/h.
Efficiency of the second tap $= 60 / 4 = 15$ units/h.
Efficiency of the third tap (emptying) $= - (60 / 5) = -12$ units/h.
When all three taps are opened simultaneously,the net efficiency $= 20 + 15 - 12 = 23$ units/h.
Time taken to fill the cistern $= \text{Total capacity} / \text{Net efficiency} = 60 / 23 \, h = 2 \frac{14}{23} \, h$.

(C) Let the total time taken to fill the tank be $x$ hours after $10$ $am$.
Pipe $A$ works for $x$ hours,and pipe $B$ works for $(x-1)$ hours (since it starts at $11$ $am$).
The rate of pipe $A$ is $1/2$ tank per hour,and the rate of pipe $B$ is $1/6$ tank per hour.
According to the problem: $\frac{x}{2} + \frac{x-1}{6} = 1$.
Multiplying the entire equation by $6$: $3x + (x - 1) = 6$.
$4x - 1 = 6 \Rightarrow 4x = 7$.
$x = 7/4$ hours,which is $1$ hour and $45$ minutes.
Since pipe $A$ started at $10$ $am$,the tank will be filled at $10$ $am + 1$ hour $45$ minutes = $11:45$ $am$.

(C) Pipe $A$ fills $\frac{1}{4}$ of the tank in $1$ $h$.
Pipe $B$ fills $\frac{1}{6}$ of the tank in $1$ $h$.
In a cycle of $2$ $h$ (opening $A$ then $B$),the part of the tank filled is $\frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$.
In $4$ $h$ (two cycles),the part of the tank filled is $2 \times \frac{5}{12} = \frac{10}{12} = \frac{5}{6}$.
Remaining part of the tank $= 1 - \frac{5}{6} = \frac{1}{6}$.
Now,it is pipe $A$'s turn to fill the remaining $\frac{1}{6}$ part.
Time taken by pipe $A$ to fill $\frac{1}{6}$ part $= \frac{1/6}{1/4} = \frac{1}{6} \times 4 = \frac{2}{3} \text{ h}$.
Total time taken $= 4 \text{ h} + \frac{2}{3} \text{ h} = 4 \frac{2}{3} \text{ h}$.

(B) Let the filling pipe be $A$ and the emptying pipe be $B$.
Rate of filling by pipe $A = \frac{1}{8}$ part per hour.
Rate of emptying by pipe $B = \frac{1}{5}$ part per hour.
When both pipes are opened,the net rate of emptying = $\frac{1}{5} - \frac{1}{8} = \frac{8-5}{40} = \frac{3}{40}$ part per hour.
Since the cistern is already $\frac{3}{4}$ full,we need to empty $\frac{3}{4}$ of the total capacity.
Time taken to empty $\frac{3}{4}$ part = $\frac{\text{Part to be emptied}}{\text{Net rate of emptying}} = \frac{3/4}{3/40} = \frac{3}{4} \times \frac{40}{3} = 10 \, h$.

(B) Let the total capacity of the tank be the Least Common Multiple $(LCM)$ of $10, 12,$ and $6$, which is $60$ units.
Efficiency of pipe $P = 60/10 = 6$ units/$h$.
Efficiency of pipe $Q = 60/12 = 5$ units/$h$.
Efficiency of pipe $C = -60/6 = -10$ units/$h$ (since it empties the tank).
Net efficiency when all three pipes are open $= 6 + 5 - 10 = 1$ unit/$h$.
We need to fill one-fourth of the tank, which is $(1/4) \times 60 = 15$ units.
Time required $= 15 \text{ units} / 1 \text{ unit}/h = 15$ hours.
Starting time is $7$ $am$.
$15$ hours after $7$ $am$ is $7$ $am + 12$ hours $= 7$ $pm$, and $7$ $pm + 3$ hours $= 10$ $pm$.
Therefore, the tank will be one-fourth full at $10$ $pm$.

(D) The rate of one tap is $\frac{1}{6}$ tank per hour.
Time taken to fill half the tank by one tap $= 6 \times \frac{1}{2} = 3$ $h$.
Remaining half of the tank is to be filled by $4$ taps (the original one plus $3$ new ones).
The combined rate of $4$ taps $= 4 \times \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$ tank per hour.
Time taken to fill the remaining half tank $= \frac{1/2}{2/3} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$ $h$.
Converting $\frac{3}{4}$ $h$ to minutes: $\frac{3}{4} \times 60 = 45$ $min$.
Total time taken $= 3$ $h + 45$ $min = 3$ $h$ $45$ $min$.

(C) Let the capacity of the tank be $V$ gallons.
Rate of pipe $A = \frac{V}{10}$ gallons/min.
Rate of pipe $B = \frac{V}{12}$ gallons/min.
Rate of pipe $C = -6$ gallons/min (since it empties the tank).
When $A, B,$ and $C$ are opened together,the net rate is $\frac{V}{20}$ gallons/min.
Therefore,the equation is: $\frac{V}{10} + \frac{V}{12} - 6 = \frac{V}{20}$.
Rearranging the terms: $\frac{V}{10} + \frac{V}{12} - \frac{V}{20} = 6$.
Taking the least common multiple $(LCM)$ of $10, 12,$ and $20$,which is $60$:
$\frac{6V + 5V - 3V}{60} = 6$.
$\frac{8V}{60} = 6$.
$\frac{2V}{15} = 6$.
$V = \frac{6 \times 15}{2} = 45$ gallons.
Thus,the capacity of the tank is $45$ gallons.

(D) Let the capacity of the cistern be the least common multiple of $40$ and $60$, which is $120$ units.
Efficiency of the first tap (filling) = $120 / 40 = 3$ units per $min$.
Efficiency of the second tap (emptying) = $120 / 60 = 2$ units per $min$.
When both taps are open, the net filling rate = $3 - 2 = 1$ unit per $min$.
Time taken to fill the cistern = $\text{Total Capacity} / \text{Net Rate} = 120 / 1 = 120$ $min$.

(B) The rate of pipe $A$ is $\frac{1}{6}$ of the cistern per hour,and the rate of pipe $B$ is $\frac{1}{8}$ of the cistern per hour.
Both pipes together fill $\left(\frac{1}{6} + \frac{1}{8}\right) = \frac{4+3}{24} = \frac{7}{24}$ of the cistern per hour.
In $2 \, h$,the part filled by both pipes is $2 \times \frac{7}{24} = \frac{14}{24} = \frac{7}{12}$.
The remaining part to be filled is $1 - \frac{7}{12} = \frac{5}{12}$.
Pipe $B$ fills the remaining part at a rate of $\frac{1}{8}$ per hour.
Time taken by $B$ to fill the remaining part $= \frac{5/12}{1/8} = \frac{5}{12} \times 8 = \frac{40}{12} = \frac{10}{3} = 3 \frac{1}{3} \, h$.

(B) Efficiency of pipe $P = \frac{1}{5}$ tank per hour.
Efficiency of pipe $Q = \frac{1}{6}$ tank per hour.
Efficiency of pipe $C = -\frac{1}{3}$ tank per hour (since it empties).
Net efficiency when all three pipes are opened $= \frac{1}{5} + \frac{1}{6} - \frac{1}{3} = \frac{6 + 5 - 10}{30} = \frac{1}{30}$ tank per hour.
To fill $\frac{2}{5}$ of the tank, the time required $= \frac{2/5}{1/30} = \frac{2}{5} \times 30 = 12 \, h$.
Starting time is $7 \, am$. Adding $12 \, h$ to $7 \, am$ gives $7 \, pm$.

(B) Let the capacity of the cistern be the Least Common Multiple $(LCM)$ of $2$ and $3$, which is $6$ units.
Efficiency of the first tap (filling) $= 6 / 2 = 3$ units per hour.
Efficiency of the second tap (emptying) $= 6 / 3 = 2$ units per hour.
When both taps are opened, the net efficiency $= 3 - 2 = 1$ unit per hour.
Time taken to fill the cistern $= \text{Total capacity} / \text{Net efficiency} = 6 / 1 = 6$ hours.

(B) Let the faster pipe take $x$ hours to fill the tank.
Then,the slower pipe takes $(x + 5)$ hours to fill the tank.
The rate of the faster pipe is $\frac{1}{x}$ tank per hour.
The rate of the slower pipe is $\frac{1}{x+5}$ tank per hour.
Working together,their combined rate is $\frac{1}{x} + \frac{1}{x+5} = \frac{1}{6}$.
Multiplying by $6x(x+5)$,we get $6(x+5) + 6x = x(x+5)$.
$6x + 30 + 6x = x^2 + 5x$.
$x^2 - 7x - 30 = 0$.
$(x - 10)(x + 3) = 0$.
Since time cannot be negative,$x = 10$.
Therefore,the faster pipe takes $10$ hours to fill the reservoir.

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

(B) Let the capacity of the cistern be $1$ unit.
Rate of first tap $= \frac{1}{12}$ units/min.
Rate of second tap $= \frac{1}{15}$ units/min.
Let the rate of the waste pipe be $x$ units/min.
When all pipes are open,the net rate is $\frac{1}{20}$ units/min.
Therefore,$\frac{1}{12} + \frac{1}{15} - x = \frac{1}{20}$.
$\frac{5+4}{60} - x = \frac{1}{20}$.
$\frac{9}{60} - x = \frac{3}{60}$.
$x = \frac{9-3}{60} = \frac{6}{60} = \frac{1}{10}$ units/min.
The waste pipe will empty the full cistern in $10$ minutes.

(C) Let the capacity of the tank be $1$ unit.
Rate of the first tap $= \frac{1}{20}$ units per minute.
Rate of the second tap $= \frac{1}{25}$ units per minute.
Combined rate of both taps $= \frac{1}{20} + \frac{1}{25} = \frac{5+4}{100} = \frac{9}{100}$ units per minute.
In $5$ minutes,the part of the tank filled by both taps $= 5 \times \frac{9}{100} = \frac{45}{100} = \frac{9}{20}$ units.
Remaining part of the tank to be filled $= 1 - \frac{9}{20} = \frac{11}{20}$ units.
Since the second tap is turned off,the first tap fills the remaining part.
Time taken by the first tap to fill $\frac{11}{20}$ units $= \frac{11/20}{1/20} = 11$ minutes.

(D) Part of the tanker filled by pipe $A$ in $1$ minute $= \frac{1}{60}$.
Part of the tanker filled by pipe $B$ in $1$ minute $= \frac{1}{40}$.
Let the total time taken to fill the tank be $2x$ minutes.
According to the problem,pipe $B$ works alone for $x$ minutes,and pipes $A$ and $B$ work together for $x$ minutes.
Therefore,the total work done is: $\frac{1}{40} \times x + (\frac{1}{60} + \frac{1}{40}) \times x = 1$.
Simplifying the equation: $\frac{x}{40} + (\frac{2+3}{120})x = 1$.
$\frac{x}{40} + \frac{5x}{120} = 1$.
$\frac{x}{40} + \frac{x}{24} = 1$.
Taking the least common multiple $(120)$: $\frac{3x + 5x}{120} = 1$.
$\frac{8x}{120} = 1$.
$x = \frac{120}{8} = 15$ minutes.
Total time taken $= 2x = 2 \times 15 = 30$ minutes.

(C) Let the capacity of the tank be $C$ gallons.
The rate of the first pipe is $\frac{C}{20}$ gallons per minute.
The rate of the second pipe is $\frac{C}{24}$ gallons per minute.
The rate of the waste pipe is $3$ gallons per minute.
When all three pipes work together,the net rate is $\frac{C}{15}$ gallons per minute.
Therefore,the equation is:
$\frac{C}{20} + \frac{C}{24} - 3 = \frac{C}{15}$
Rearranging the terms to solve for $C$:
$\frac{C}{20} + \frac{C}{24} - \frac{C}{15} = 3$
Finding a common denominator $(120)$:
$\frac{6C + 5C - 8C}{120} = 3$
$\frac{3C}{120} = 3$
$\frac{C}{40} = 3$
$C = 120$ gallons.
Thus,the capacity of the tank is $120$ gallons.

(D) Part of the tank filled by pipe $A$ and $B$ in $1$ minute $= \frac{1}{15} + \frac{1}{20} = \frac{4+3}{60} = \frac{7}{60}$.
In the first $4$ minutes,the part of the tank filled $= 4 \times \frac{7}{60} = \frac{28}{60} = \frac{7}{15}$.
Remaining part of the tank to be filled $= 1 - \frac{7}{15} = \frac{8}{15}$.
Since pipe $A$ is turned off,the remaining part is filled by pipe $B$ alone.
Let the time taken by pipe $B$ to fill the remaining part be $x$ minutes.
$\frac{1}{20} \times x = \frac{8}{15} \implies x = \frac{8 \times 20}{15} = \frac{8 \times 4}{3} = \frac{32}{3} = 10 \frac{2}{3}$ minutes.
$10 \frac{2}{3}$ minutes $= 10$ minutes and $\frac{2}{3} \times 60$ seconds $= 10$ minutes $40$ seconds.
Total time taken to fill the tank $= 4$ minutes $+ 10$ minutes $40$ seconds $= 14$ minutes $40$ seconds.

(C) Let the work done by pipes $A, B,$ and $C$ in $1$ hour be $\frac{1}{A}, \frac{1}{B},$ and $\frac{1}{C}$ respectively.
Given that $(A+B+C)$ can fill the tank in $6$ hours,so $(A+B+C)$'s $1$ hour work $= \frac{1}{6}$.
Work done by $A, B,$ and $C$ in $2$ hours $= 2 \times \frac{1}{6} = \frac{1}{3}$.
Remaining part of the tank $= 1 - \frac{1}{3} = \frac{2}{3}$.
This remaining part is filled by $A$ and $B$ in $7$ hours,so $(A+B)$'s $1$ hour work $= \frac{2/3}{7} = \frac{2}{21}$.
Now,$C$'s $1$ hour work $= (A+B+C)$'s $1$ hour work $- (A+B)$'s $1$ hour work.
$C$'s $1$ hour work $= \frac{1}{6} - \frac{2}{21} = \frac{7 - 4}{42} = \frac{3}{42} = \frac{1}{14}$.
Therefore,$C$ alone can fill the tank in $14$ hours.

(C) Let the time taken by the second pipe to fill the tank be $x$ hours.
Then,the time taken by the first pipe is $x+5$ hours,and the time taken by the third pipe is $x-4$ hours.
According to the problem,the first two pipes working together fill the tank in the same time as the third pipe alone.
Therefore,the rate of the first pipe plus the rate of the second pipe equals the rate of the third pipe:
$\frac{1}{x+5} + \frac{1}{x} = \frac{1}{x-4}$
$\frac{x + (x+5)}{x(x+5)} = \frac{1}{x-4}$
$\frac{2x+5}{x^2+5x} = \frac{1}{x-4}$
$(2x+5)(x-4) = x^2+5x$
$2x^2 - 8x + 5x - 20 = x^2 + 5x$
$2x^2 - 3x - 20 = x^2 + 5x$
$x^2 - 8x - 20 = 0$
$(x-10)(x+2) = 0$
Since time cannot be negative,$x = 10$ hours.
The time required by the first pipe is $x+5 = 10+5 = 15$ hours.