Calculating Pipe Rates
First, determine the filling/draining rate for each pipe:
- Pipe A fills the cistern in 18 hours. Rate (A) = $\frac{1}{18}$ cistern/hour.
- Pipe B fills the cistern in 27 hours. Rate (B) = $\frac{1}{27}$ cistern/hour.
- Pipe C drains the cistern in 45 hours. Rate (C) = $-\frac{1}{45}$ cistern/hour (negative sign indicates draining).
Analyzing the Filling Process
The process occurs in two phases:
- Phase 1: Pipes A and C are turned on. Let this phase last for $t_1$ hours.
- Phase 2: Pipe A is turned off, and Pipe B is turned on instantly. Pipe C remains on. Let this phase last for $t_2$ hours.
The total time to fill the cistern is given as 55 hours. Therefore:
$t_1 + t_2 = 55 \text{ hours}$
Setting Up the Work Equation
The total work done is filling 1 cistern. The work done is the sum of the work done by each pipe in its respective phase.
Combined rate in Phase 1 (A + C) = Rate(A) + Rate(C) = $\frac{1}{18} - \frac{1}{45} = \frac{5 - 2}{90} = \frac{3}{90} = \frac{1}{30}$ cistern/hour.
Combined rate in Phase 2 (B + C) = Rate(B) + Rate(C) = $\frac{1}{27} - \frac{1}{45} = \frac{5 - 3}{135} = \frac{2}{135}$ cistern/hour.
The total work equation is:
$(\text{Rate A + Rate C}) \times t_1 + (\text{Rate B + Rate C}) \times t_2 = 1$
$ \left( \frac{1}{30} \right) t_1 + \left( \frac{2}{135} \right) t_2 = 1 $
Solving for Pipe B's Duration ($t_2$)
We know $t_1 = 55 - t_2$. Substitute this into the work equation:
$ \frac{1}{30} (55 - t_2) + \frac{2}{135} t_2 = 1 $
Multiply the equation by the Least Common Multiple (LCM) of the denominators (30 and 135), which is 270:
$ 270 \times \frac{1}{30} (55 - t_2) + 270 \times \frac{2}{135} t_2 = 270 \times 1 $
$ 9 (55 - t_2) + 2 \times 2 t_2 = 270 $
$ 495 - 9 t_2 + 4 t_2 = 270 $
$ 495 - 5 t_2 = 270 $
Rearrange to solve for $t_2$:
$ 5 t_2 = 495 - 270 $
$ 5 t_2 = 225 $
$ t_2 = \frac{225}{5} $
$ t_2 = 45 \text{ hours} $
Pipe B was turned on for 45 hours.
Similar Questions
- Three pipes of diameters 2 cm, 3 cm and 4 cm are running together to fill a cistern. The smallest pipe alone can fill the cistern in 232 minutes. The amount of water flowing in each pipe is proportional to the square of its diameter. The time (in minutes) taken by all pipes running together to completely fill the cistern is:
- A pipe can fill a sump with water in 2 hours. Because of a leak, it took $2 \frac{1}{3}$ hours to fill the sump. The leak can drain all the water of the sump in:
- Pipes A, B and C are attached to an empty cistern. While the first two can fill the cistern in 4 and 10 hours, respectively, the third can drain the cistern, when filled, in 6 hours. If all the three pipes are opened simultaneously when the cistern is half-full, how many hours will be needed to fill the cistern?
Important Questions from Pipe and Cistern
A pipe can fill a cistern in 20 minutes where as the cistern when full can be emptied by a leak in 28 minutes. When both are opened, The time taken to fill the cistern is:
‘A’ pipe can empty a tank in 20 minutes. The second pipe ‘B’ has a diameter twice as that of ‘A’. If both A & B pipe are attached to the tank how much time will be required to empty the tank?
Pipes A and B can empty a full tank in 16 hours and 24 hours, respectively. Pipe C alone can fill the empty tank in 4 hours. If A, B and C are opened together, the tank will be 35% full after :
Pipes A and B can fill a tank in 36 minutes and 45 minutes, respectively. Both these pipes were opened simultaneously. After 20 minutes, a leak at the bottom of the tank was spotted which was immediately sealed. The tank was full in another 15 minutes. The leak alone can empty the full tank in:
A cistern has a leak which would empty it in 6 hours. A tap is turned on which admits 10 litres of water per minute into the cistern. When it is full it is now emptied in 10 hours. What is the capacity (in litres) of the cistern?
