Understanding Pipe Filling Rates
First, let's determine the rate at which each pipe fills the tank. The rate is the fraction of the tank filled per hour.
- Rate of Pipe A = $ \frac{1}{8} $ tank per hour
- Rate of Pipe B = $ \frac{1}{12} $ tank per hour
- Rate of Pipe C = $ \frac{1}{24} $ tank per hour
Calculating Work Done in Stages
The filling process occurs in three distinct stages.
Stage 1: Pipe A Alone (1 hour)
Pipe A works alone for the first hour.
- Work done by A in 1 hour = Rate of A $ \times $ Time = $ \frac{1}{8} \times 1 = \frac{1}{8} $ tank.
- Tank capacity remaining = $ 1 - \frac{1}{8} = \frac{7}{8} $ tank.
Stage 2: Pipes A and B Together (2 hours)
Next, pipes A and B work together for 2 hours.
- Combined rate of A and B = $ \frac{1}{8} + \frac{1}{12} $.
- To add these fractions, find a common denominator (24): $ \frac{3}{24} + \frac{2}{24} = \frac{5}{24} $ tank per hour.
- Work done by A and B in 2 hours = Combined rate $ \times $ Time = $ \frac{5}{24} \times 2 = \frac{10}{24} = \frac{5}{12} $ tank.
- Total work done after 2 hours = Work from Stage 1 + Work from Stage 2 = $ \frac{1}{8} + \frac{5}{12} $.
- Common denominator (24): $ \frac{3}{24} + \frac{10}{24} = \frac{13}{24} $ tank.
- Tank capacity remaining = $ 1 - \frac{13}{24} = \frac{11}{24} $ tank.
Stage 3: Pipes A, B, and C Together
Finally, all three pipes work together until the tank is full.
- Combined rate of A, B, and C = $ \frac{1}{8} + \frac{1}{12} + \frac{1}{24} $.
- Common denominator (24): $ \frac{3}{24} + \frac{2}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4} $ tank per hour.
Determining Final Filling Time
Now, calculate the time needed for Stage 3 and the total time.
- Time required for Stage 3 = Remaining capacity / Combined rate of A, B, C
- Time for Stage 3 = $ \frac{11/24}{1/4} = \frac{11}{24} \times 4 = \frac{44}{24} = \frac{11}{6} $ hours.
- Total time to fill the tank = Time (Stage 1) + Time (Stage 2) + Time (Stage 3)
- Total time = $ 1 \text{ hour} + 2 \text{ hours} + \frac{11}{6} \text{ hours} = 3 + \frac{11}{6} $ hours.
- Total time = $ \frac{18}{6} + \frac{11}{6} = \frac{29}{6} $ hours.
Convert the total time into hours and minutes:
- $ \frac{29}{6} $ hours = 4 whole hours and $ \frac{5}{6} $ of an hour.
- $ \frac{5}{6} $ hour = $ \frac{5}{6} \times 60 $ minutes = 50 minutes.
- Therefore, the total time is 4 hours and 50 minutes.
Important Questions from Pipe and Cistern
Pipes A, B and C can fill a tank in 20, 30 and 60 hours, respectively. Pipes A, B and C are opened at 7 a.m., 8 a.m., and 9 a.m., respectively, on the same day. When will the tank be full?
There are two water taps in a tank which can fill the empty tank in 12 hours and 18 hours respectively. It is seen that there is a leakage point at the bottom of the tank which can empty the completely filled tank in 36 hours. If both the water taps are opened at the same time to fill the empty tank and leakage point was repaired after 1 hour, then in how much time the empty tank will be completely filled?
Two pipes A and B can fill a tank in 12 minutes and 24 minutes, respectively, while a third pipe C can empty the full tank in 32 minutes. All the three pipes are opened simultaneously. However, pipe C is closed 2 minutes before the tank is filled. In how much time (in minutes) will the tank be full?
Pipes A and B can fill a tank in 12 hours and 16 hours respectively and pipe C can empty the full tank in 24 hours. All three pipes are opened together, but after 4 hours pipe B is closed. In how many hours, the empty tank will be completely filled?
Pipes A and B can fill a tank in 43.2 minutes and 108 minutes, respectively. Pipe C can empty it at 3 litres/minute. When all the three pipes are opened together, they fill the tank in 54 minutes. The capacity (in litres) of the tank is:
