This problem involves calculating the time required to fill a partially filled cistern when multiple pipes with different functions (filling and draining) are operating simultaneously.
Pipe Rates Calculation
First, determine the rate at which each pipe works. The rate is the fraction of the cistern filled or drained per hour.
- Pipe A fills the cistern in 4 hours. Rate of Pipe A = $ \frac{1}{4} $ cistern/hour.
- Pipe B fills the cistern in 10 hours. Rate of Pipe B = $ \frac{1}{10} $ cistern/hour.
- Pipe C drains the cistern in 6 hours. Rate of Pipe C = $ -\frac{1}{6} $ cistern/hour (negative because it drains).
Combined Rate Calculation
When all three pipes are opened simultaneously, their rates add up. Calculate the net rate of filling:
Combined Rate = Rate(A) + Rate(B) + Rate(C)
Combined Rate = $ \frac{1}{4} + \frac{1}{10} - \frac{1}{6} $
To add/subtract these fractions, find a common denominator. The Least Common Multiple (LCM) of 4, 10, and 6 is 60.
Combined Rate = $ \frac{1 \times 15}{4 \times 15} + \frac{1 \times 6}{10 \times 6} - \frac{1 \times 10}{6 \times 10} $
Combined Rate = $ \frac{15}{60} + \frac{6}{60} - \frac{10}{60} $
Combined Rate = $ \frac{15 + 6 - 10}{60} = \frac{11}{60} $ cistern/hour.
This means that when all pipes are open, the cistern fills at a net rate of $ \frac{11}{60} $ of its capacity every hour.
Time Calculation for Half-Full Cistern
The cistern is initially half-full. This means only half of the cistern needs to be filled.
Work to be done = $ \frac{1}{2} $ cistern.
The formula relating time, work, and rate is: Time = Work / Rate.
Time = $ \frac{\text{Work to be done}}{\text{Combined Rate}} $
Time = $ \frac{1/2}{11/60} $
Time = $ \frac{1}{2} \times \frac{60}{11} $
Time = $ \frac{60}{22} = \frac{30}{11} $ hours.
Therefore, it will take $ \frac{30}{11} $ hours to fill the remaining half of the cistern.
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:
- Two pipes A and B can fill an empty cistern in 18 and 27 hours, respectively. Pipe C can drain the entire cistern in 45 hours when no other pipe is in operation. Initially, when the cistern was empty Pipe A and Pipe C were turned on. After a few hours Pipe A was turned off and Pipe B was turned on instantly. In all, it took 55 hours to fill the cistern. For how many hours was Pipe B turned on?
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?
