Introduction

Suppose we have two pipes, Pipe A and Pipe B, that can independently fill a tank. Pipe A can fill the tank in 30 minutes while Pipe B can do so in 20 minutes. The initial problem is to determine the exact time after which Pipe B should be switched off so that both pipes, when working together, can fill the tank in exactly 24 minutes. This article will detail the mathematical steps to solve such a problem.

Mechanics of Filling the Tank

To solve the problem, let's first determine the rates at which each pipe fills the tank.

Rate of Pipe A: Since Pipe A can fill the tank in 30 minutes, its rate is :

[ text{Rate of Pipe A} frac{1}{30} text{ tanks per minute} ]

Rate of Pipe B: Since Pipe B can fill the tank in 20 minutes, its rate is :

[ text{Rate of Pipe B} frac{1}{20} text{ tanks per minute} ]

The combined rate at which both pipes fill the tank when they are open simultaneously can be calculated as follows:

Combined Rate: The least common multiple of 30 and 20 is 60, hence:

[ text{Rate} frac{2}{60} frac{3}{60} frac{5}{60} frac{1}{12} text{ tanks per minute} ]

Given that, in 24 minutes, the combined rate would fill:

[ text{Total Work} frac{1}{12} times 24 2 text{ tanks} ]

However, we need to fill only 1 tank exactly in 24 minutes. To achieve this, we need to determine the time during which Pipe B should be on.

Determining the Time for Pipe B to Stay On

Let's denote the time during which both pipes are open simultaneously as ( t ). During this time, they fill:

[ text{Work done} frac{1}{12} times t ]

After ( t ) minutes, Pipe B is turned off, and only Pipe A continues to work for the remaining ( 24-t ) minutes, filling:

[ text{Additional Work} frac{1}{30} times (24 - t) ]

The total work done must equal 1 tank:

[ frac{1}{12}t frac{1}{30}(24 - t) 1 ]

Combining the fractions, the common denominator is 60:

[ frac{5t}{60} frac{224 - 2t}{60} 1 ]

Which simplifies to:

[ 5t 224 - 2t 60 ]

Combining like terms gives:

[ 3t 224 60 ]

Subtracting 224 from both sides:

[ 3t 48 ]

Solving for ( t ) gives:

[ t 16 ]

Conclusion

Therefore, Pipe B should be switched off after 4 minutes. During the initial 4 minutes, both pipes work together to fill the tank, and then Pipe A alone works for the remaining 20 minutes, resulting in the tank being filled exactly in 24 minutes.

This problem demonstrates the application of simple algebra and rate theory in real-world scenarios. It is a practical tool for optimizing resource usage and understanding the interplay between different entities working towards a common goal.

Keywords

Pipe scheduling, tank filling, mathematical optimization