Filling a Tank with Two Pipes and a Leak: A Comprehensive Solution

In this article, we will explore a practical problem involving two pipes filling a tank and a leak that develops. The problem involves a systematic step-by-step approach to determine the time taken to fill the tank under these conditions. We will break down the problem into manageable steps and provide a detailed solution using mathematical reasoning and calculations.

Introduction

This problem is a classic example of a rate and time calculation, which is a common topic in mathematics and engineering. Understanding how to approach and solve such problems is essential for anyone dealing with real-world scenarios where multiple factors come into play, such as flow rates, leaks, and efficient use of resources.

Problem Statement

Two pipes can fill a tank in 20 hours and 30 hours, respectively. Both pipes are opened, but when the tank is 1/3 full, a leak develops that allows 1/3 of the water supplied by both pipes to leak out. We need to determine the time taken to fill the tank in these conditions.

Step-by-Step Solution

Step 1: Determine the Rates of the Pipes

The first step is to calculate the rates at which the pipes fill the tank. We start by determining the rate of each pipe individually.

For Pipe A:

Rate of Pipe A  frac{1}{20} text{ tanks per hour}

For Pipe B:

Rate of Pipe B  frac{1}{30} text{ tanks per hour}

Next, we find the combined rate of both pipes:

Combined Rate  frac{1}{20}   frac{1}{30}

To add these fractions, we need a common denominator, which is 60:

Combined Rate  frac{3}{60}   frac{2}{60}  frac{5}{60}  frac{1}{12} text{ tanks per hour}

Step 2: Calculate the Time to Fill 1/3 of the Tank

Initially, both pipes are opened, and we need to determine how long it takes to fill 1/3 of the tank.

Time to fill frac{1}{3} text{ tank}  frac{frac{1}{3}}{frac{1}{12}}  frac{1}{3} times 12  4 text{ hours}

Here, we used the combined rate of both pipes to find the time taken to fill 1/3 of the tank.

Step 3: Determine the Volume Filled When the Leak Starts

Upon reaching 1/3 of the tank, a leak develops that allows 1/3 of the water supplied by both pipes to leak out. We calculate the volume filled in 4 hours using the combined rate.

Volume filled  frac{1}{12} times 4  frac{1}{3} text{ tank}

This means that after 4 hours, 1/3 of the tank is filled.

Step 4: Calculate the Effect of the Leak

With the leak developing, we need to calculate the net water supplied by both pipes in 1 hour after the leak starts.

Water supplied in 1 hour by both pipes  frac{1}{12} text{ tank}

The leak allows 1/3 of this water to leak out:

Water lost  frac{1}{3} times frac{1}{12}  frac{1}{36} text{ tank}

Therefore, the net water supplied in 1 hour is:

Net supply  frac{1}{12} - frac{1}{36}

We find a common denominator of 36 to subtract these fractions:

Net supply  frac{3}{36} - frac{1}{36}  frac{2}{36}  frac{1}{18} text{ tanks per hour}

Step 5: Calculate the Remaining Volume to Fill

The tank is currently 1/3 full, so the remaining volume is:

Remaining volume  1 - frac{1}{3}  frac{2}{3} text{ tank}

Step 6: Calculate the Time to Fill the Remaining Volume

Now we can find the time taken to fill the remaining 2/3 of the tank at the net supply rate:

Time to fill frac{2}{3} text{ tank}  frac{frac{2}{3}}{frac{1}{18}}  frac{2}{3} times 18  12 text{ hours}

This calculation shows the time required to fill the remaining part of the tank.

Step 7: Calculate the Total Time Taken to Fill the Tank

Finally, we add the time taken to fill the first 1/3 of the tank and the time taken to fill the remaining 2/3:

Total time  4 text{ hours}   12 text{ hours}  16 text{ hours}

Therefore, the total time taken to fill the tank is 16 hours.

Conclusion

This problem provides a clear example of how to handle rate, time, and volume calculations, especially when a leak is involved. Understanding these concepts and solving such problems is crucial in various applications, such as piping systems, water supply networks, and leak detection and repair. The step-by-step approach used here can be applied to similar scenarios to find solutions efficiently.