Optimizing Cistern Filling with Pipes A, B, and C

In this article, we will delve into a practical problem involving three pipes, A, B, and C, used to fill a cistern. Understanding pipe rates, calculating the net effect of open and closed periods, and using these techniques to optimize the filling process are the main topics we will explore. This is not just a theoretical exercise but a useful skill in various real-world scenarios, such as industrial processes or water management systems. By employing these methods, you can achieve efficient cistern filling, reducing time and resources.

Understanding Pipe Rates

To begin, we must determine the rates at which pipes A, B, and C fill or empty the cistern. This is crucial for understanding the overall process and making informed decisions on when to open or close the pipes.

Rate of Pipe A

Pipe A can fill the cistern in 20 minutes. Therefore, the rate of Pipe A is:

(frac{1}{20}) of the cistern per minute.

Rate of Pipe B

Pipe B can fill the cistern in 30 minutes. Consequently, the rate of Pipe B is:

(frac{1}{30}) of the cistern per minute.

Rate of Pipe C

Pipe C can empty the cistern in 15 minutes, making its rate:

-(frac{1}{15}) of the cistern per minute (negative because it removes water).

Step-by-Step Calculation

In this section, we will perform a step-by-step calculation to determine the net effect of opening the pipes in the order A, B, C for one minute each, then analyze how long it will take to fill the cistern.

First Minute (Pipe A)

Amount filled by Pipe A in one minute:

(frac{1}{20})

Second Minute (Pipe B)

Amount filled by Pipe B in one minute:

(frac{1}{30})

Third Minute (Pipe C)

Amount emptied by Pipe C in one minute:

-(frac{1}{15})

Total amount filled after three minutes:

(left(frac{1}{20} frac{1}{30} - frac{1}{15}right))

Calculating the Total Amount Filled

Since the denominators are 20, 30, and 15, the least common multiple (LCM) is 60. Converting the fractions:

(frac{1}{20} frac{3}{60})

(frac{1}{30} frac{2}{60})

( frac{1}{15} frac{4}{60})

So the total filled amount in three minutes is:

(left(frac{3}{60} frac{2}{60} - frac{4}{60}right) frac{1}{60})

This means that in three minutes, the cistern is filled by (frac{1}{60}) of its capacity.

Time to Fill the Cistern Completingly

Given that (frac{1}{60}) of the cistern is filled in 3 minutes, the amount filled in one minute is:

(frac{1}{60} div 3 frac{1}{180})

To fill the entire cistern, we calculate the total time as follows:

(60 text{ full cistern} times 3 text{ minutes per cycle} 180 text{ minutes})

In the first 3 minutes, the cistern has been filled by (frac{1}{60}). We need to determine how much more is required to reach full capacity:

(1 - frac{1}{60} frac{59}{60})

Calculating Additional Cycles

Since each cycle fills (frac{1}{60}) of the cistern, we can find out how many complete cycles are needed to fill (frac{59}{60}) of the cistern:

( frac{59}{60} div frac{1}{60} 59 text{ cycles})

Calculating the total time for these cycles:

(59 text{ cycles} times 3 text{ minutes per cycle} 177 text{ minutes})

Adding the initial 3 minutes:

(177 3 180 text{ minutes})

Thus, the total time taken to fill the cistern completely is 180 minutes.

Conclusion

By systematically calculating the rates and understanding the net effect of opening the pipes in a specific sequence, we can optimize the process of filling the cistern. This method is not only educational but also practical, helping to minimize the time and resources required in similar scenarios.