Pipes and Cistern Questions in English

(C) Pipe $A$ fills the tank in $2$ hours,so in $1$ hour,it fills $\frac{1}{2}$ of the tank.
Pipe $B$ fills the tank in $6$ hours,so in $1$ hour,it fills $\frac{1}{6}$ of the tank.
Pipe $A$ is opened at $10 \, am$. By $11 \, am$,it has worked for $1$ hour and filled $\frac{1}{2}$ of the tank.
Remaining part of the tank to be filled $= 1 - \frac{1}{2} = \frac{1}{2}$.
At $11 \, am$,both pipes $A$ and $B$ are open. Their combined efficiency per hour $= \frac{1}{2} + \frac{1}{6} = \frac{3+1}{6} = \frac{4}{6} = \frac{2}{3}$ of the tank per hour.
Time taken to fill the remaining $\frac{1}{2}$ part $= \frac{\text{Remaining Part}}{\text{Combined Efficiency}} = \frac{1/2}{2/3} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$ hours.
$\frac{3}{4}$ hours $= \frac{3}{4} \times 60 = 45$ minutes.
Therefore,the tank will be filled at $11 \, am + 45$ minutes $= 11:45 \, am$.

(A) Let the time taken by pipe $A$ be $x$ hours. Since pipe $A$ takes half the time of pipe $B$,the time taken by pipe $B$ is $2x$ hours.
The combined rate of pipes $A$ and $B$ is $\frac{1}{x} + \frac{1}{2x} = \frac{3}{2x}$.
According to the problem,this combined rate is twice the rate of pipe $C$ (since they take half the time of $C$). Therefore,the rate of pipe $C$ is $\frac{1}{2} \times \frac{3}{2x} = \frac{3}{4x}$.
Together,all three pipes take $6$ hours $40$ minutes,which is $6 + \frac{40}{60} = 6 + \frac{2}{3} = \frac{20}{3}$ hours.
The combined rate of all three pipes is $\frac{1}{x} + \frac{1}{2x} + \frac{3}{4x} = \frac{1}{20/3} = \frac{3}{20}$.
Solving for $x$: $\frac{4+2+3}{4x} = \frac{3}{20} \Rightarrow \frac{9}{4x} = \frac{3}{20}$.
$12x = 180 \Rightarrow x = 15$ hours.

(D) Let the total capacity of the tank be $1$ unit.
Rate of filling by tap $A = \frac{1}{4}$ unit per hour.
Rate of filling by tap $B = \frac{1}{6}$ unit per hour.
Rate of emptying by tap $C = \frac{1}{3}$ unit per hour.
When all three taps are opened simultaneously,the net rate of filling is:
$= \left(\frac{1}{4} + \frac{1}{6} - \frac{1}{3}\right)$ unit per hour.
$= \left(\frac{3 + 2 - 4}{12}\right) = \frac{1}{12}$ unit per hour.
Therefore,the time required to fill the tank completely is the reciprocal of the net rate:
$= \frac{1}{1/12} = 12$ $hours$.

(A) Let the total capacity of the cistern be the $LCM$ of $3, 4,$ and $1$,which is $12$ units.
Efficiency of pipe $A = +4$ units/hour.
Efficiency of pipe $B = +3$ units/hour.
Efficiency of pipe $C = -12$ units/hour.
Pipe $A$ is opened at $3 \text{ pm}$. By $5 \text{ pm}$,it has worked for $2$ hours. Work done $= 2 \times 4 = 8$ units.
Pipe $B$ is opened at $4 \text{ pm}$. By $5 \text{ pm}$,it has worked for $1$ hour. Work done $= 1 \times 3 = 3$ units.
Total work done by $5 \text{ pm} = 8 + 3 = 11$ units.
At $5 \text{ pm}$,all three pipes are open. Net efficiency $= 4 + 3 - 12 = -5$ units/hour.
This means the cistern is being emptied at a rate of $5$ units/hour.
Time taken to empty the $11$ units $= \frac{11}{5} \text{ hours} = 2 \text{ hours and } 12 \text{ minutes}$.
Therefore,the cistern will be empty at $5 \text{ pm} + 2 \text{ hours } 12 \text{ minutes} = 7:12 \text{ pm}$.

(D) Let the time taken by the faster pipe to fill the tank be $x$ minutes.
Since the other pipe is three times slower,it will take $3x$ minutes to fill the tank.
The work done by the faster pipe in $1$ minute is $\frac{1}{x}$ and by the slower pipe is $\frac{1}{3x}$.
Given that together they fill the tank in $36$ minutes,their combined work rate is $\frac{1}{36}$.
Therefore,$\frac{1}{x} + \frac{1}{3x} = \frac{1}{36}$.
Multiplying by $3x$,we get $3 + 1 = \frac{3x}{36}$,which simplifies to $4 = \frac{x}{12}$.
Thus,$x = 48$ minutes.
The slower pipe takes $3x = 3 \times 48 = 144$ minutes.
$144$ minutes is equal to $2$ hours and $24$ minutes.

(B) Let the capacity of the tank be $C$ gallons.
The first pipe fills the tank in $24$ minutes,so its rate is $\frac{C}{24}$ gallons/minute.
The second pipe fills the tank in $40$ minutes,so its rate is $\frac{C}{40}$ gallons/minute.
The third pipe empties the tank at a rate of $30$ gallons/minute.
When all three pipes are open,the net rate of filling is $\frac{C}{24} + \frac{C}{40} - 30$ gallons/minute.
Since the tank is filled in $1$ hour ($60$ minutes),we have:
$60 \times (\frac{C}{24} + \frac{C}{40} - 30) = C$
Simplify the expression inside the parentheses:
$\frac{5C + 3C}{120} - 30 = \frac{C}{60}$
$\frac{8C}{120} - 30 = \frac{C}{60}$
$\frac{C}{15} - 30 = \frac{C}{60}$
Multiply the entire equation by $60$ to clear the denominators:
$4C - 1800 = C$
$3C = 1800$
$C = 600$
Therefore,the capacity of the tank is $600$ gallons.