Two pipes fill a tank when working individually in 25 and 40 hours, respectively while a third pipe can drain the filled tank in 16 hours. If all the three pipes are turned on at the same time when the tank is empty, how long will it take to fill the tank completely?
Understanding Pipe and Tank Filling Problems
This problem involves calculating the time taken to fill a tank when multiple pipes are working simultaneously, some filling the tank and others draining it. The key concept here is understanding the work rate of each pipe. The work rate is the amount of work done per unit of time. In this case, the "work" is filling (or draining) the tank, and the unit of time is hours.
Calculating Individual Work Rates
If a pipe can fill a tank in $T$ hours, its work rate is $\frac{1}{T}$ of the tank filled per hour. If a pipe can drain a tank in $T$ hours, its work rate is $-\frac{1}{T}$ of the tank drained per hour (negative to indicate draining).
- Pipe 1 fills the tank in 25 hours. Its filling rate is $\frac{1}{25}$ tank per hour.
- Pipe 2 fills the tank in 40 hours. Its filling rate is $\frac{1}{40}$ tank per hour.
- Pipe 3 drains the tank in 16 hours. Its draining rate is $-\frac{1}{16}$ tank per hour.
Calculating Combined Work Rate
When all three pipes are turned on at the same time, their work rates are combined. Filling rates are added, and draining rates are subtracted.
Combined rate = (Rate of Pipe 1) + (Rate of Pipe 2) + (Rate of Pipe 3)
Combined rate = $\frac{1}{25} + \frac{1}{40} - \frac{1}{16}$ tank per hour.
To add and subtract these fractions, we find a common denominator for 25, 40, and 16.
- Prime factorization of 25 is $5^2$.
- Prime factorization of 40 is $2^3 \times 5$.
- Prime factorization of 16 is $2^4$.
The least common multiple (LCM) of 25, 40, and 16 is $2^4 \times 5^2 = 16 \times 25 = 400$.
Now, we express each fraction with the denominator 400:
- $\frac{1}{25} = \frac{1 \times 16}{25 \times 16} = \frac{16}{400}$
- $\frac{1}{40} = \frac{1 \times 10}{40 \times 10} = \frac{10}{400}$
- $-\frac{1}{16} = -\frac{1 \times 25}{16 \times 25} = -\frac{25}{400}$
Combined rate = $\frac{16}{400} + \frac{10}{400} - \frac{25}{400} = \frac{16 + 10 - 25}{400} = \frac{26 - 25}{400} = \frac{1}{400}$ tank per hour.
Calculating Total Time to Fill the Tank
The total time taken to fill the tank is the reciprocal of the combined work rate.
Time = $\frac{1}{\text{Combined rate}} = \frac{1}{\frac{1}{400}}$ hours.
Time = 400 hours.
Converting Hours to Days and Hours
We need to convert 400 hours into days and hours. There are 24 hours in a day.
Divide 400 by 24:
$400 \div 24$
$400 = 24 \times 16 + 16$
This means 400 hours is equal to 16 full days and 16 remaining hours.
So, it will take 16 days and 16 hours to fill the tank completely when all three pipes are working together.
Revision Table: Pipe and Tank Calculations
| Pipe | Time (hours) | Type | Rate (tank/hour) |
|---|---|---|---|
| Pipe 1 | 25 | Filling | $\frac{1}{25}$ |
| Pipe 2 | 40 | Filling | $\frac{1}{40}$ |
| Pipe 3 | 16 | Draining | $-\frac{1}{16}$ |
| Combined | 400 | Filling (Net) | $\frac{1}{400}$ |
Additional Information: Work and Time Concepts
Problems involving pipes and tanks are a common application of work and time concepts. The fundamental principle is that the total work done is equal to the rate of work multiplied by the time taken.
In these problems:
- The "work" is often considered as completing one task, like filling or draining a tank (representing 1 unit of work).
- The "rate" is how much of the tank is filled or drained per unit of time.
- If multiple agents (pipes) are working, their rates are combined (added for agents doing the same work, subtracted for agents doing opposite work).
If a pipe's filling time is $T$, its rate is $\frac{1}{T}$. The time taken for the combined work is $\frac{1}{\text{Combined Rate}}$. This relationship is inverse; a higher rate means less time to complete the work.
Similar Questions
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