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?
Understanding the Pipes and Tank Filling Problem
This problem involves three pipes that can fill a tank at different rates and are opened at different times. To find the total time it takes for the tank to be full, we need to calculate the rate of work for each pipe, determine how much work is done in the initial hours when not all pipes are open, and then calculate the time required to finish filling the tank when all pipes are working together.
Calculating Individual Pipe Filling Rates
The rate at which a pipe fills a tank is the reciprocal of the time it takes to fill the entire tank alone. The rates are:
- Pipe A fills the tank in 20 hours. Rate of A = \( \frac{1}{20} \) of the tank per hour.
- Pipe B fills the tank in 30 hours. Rate of B = \( \frac{1}{30} \) of the tank per hour.
- Pipe C fills the tank in 60 hours. Rate of C = \( \frac{1}{60} \) of the tank per hour.
Work Done Before All Pipes Are Opened
The pipes are opened sequentially:
- Pipe A is opened at 7 a.m.
- Pipe B is opened at 8 a.m.
- Pipe C is opened at 9 a.m.
Let's calculate the work done hour by hour until all pipes are open:
- From 7 a.m. to 8 a.m. (1 hour): Only Pipe A is working.
- Work done by A in the first hour = \( 1 \text{ hour} \times \frac{1}{20} \text{ tank/hour} = \frac{1}{20} \) of the tank.
- From 8 a.m. to 9 a.m. (1 hour): Pipes A and B are working.
- Combined rate of A and B = Rate of A + Rate of B = \( \frac{1}{20} + \frac{1}{30} \).
- To add these fractions, find a common denominator, which is 60.
- Combined rate of A and B = \( \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \) of the tank per hour.
- Work done by A and B in the second hour = \( 1 \text{ hour} \times \frac{1}{12} \text{ tank/hour} = \frac{1}{12} \) of the tank.
Total work done by 9 a.m. = Work done in hour 1 + Work done in hour 2
Total work done by 9 a.m. = \( \frac{1}{20} + \frac{1}{12} \)
Using the common denominator 60:
Total work done by 9 a.m. = \( \frac{3}{60} + \frac{5}{60} = \frac{8}{60} = \frac{2}{15} \) of the tank.
Calculating the Remaining Work
The total capacity of the tank is considered as 1 unit of work. The remaining work to be done after 9 a.m. is:
Remaining Work = Total Work - Work done by 9 a.m.
Remaining Work = \( 1 - \frac{2}{15} = \frac{15}{15} - \frac{2}{15} = \frac{13}{15} \) of the tank.
Combined Filling Rate of All Pipes (A, B, and C)
From 9 a.m. onwards, all three pipes A, B, and C are working together. Their combined filling rate is the sum of their individual rates:
Combined rate of A, B, and C = Rate of A + Rate of B + Rate of C
Combined rate = \( \frac{1}{20} + \frac{1}{30} + \frac{1}{60} \)
Using the common denominator 60:
Combined rate = \( \frac{3}{60} + \frac{2}{60} + \frac{1}{60} = \frac{3+2+1}{60} = \frac{6}{60} = \frac{1}{10} \) of the tank per hour.
Time to Fill the Remaining Tank
The time taken to fill the remaining \( \frac{13}{15} \) of the tank at a combined rate of \( \frac{1}{10} \) tank/hour is calculated as:
Time = Remaining Work / Combined Rate
Time = \( \frac{\frac{13}{15}}{\frac{1}{10}} = \frac{13}{15} \times \frac{10}{1} = \frac{13 \times 10}{15} = \frac{130}{15} \) hours.
Simplify the fraction \( \frac{130}{15} \) by dividing both numerator and denominator by 5:
Time = \( \frac{26}{3} \) hours.
Now, convert this fraction of hours into hours and minutes. \( \frac{26}{3} \) hours is 8 with a remainder of 2. So, it is 8 full hours and \( \frac{2}{3} \) of an hour.
Convert the fraction of an hour to minutes:
\( \frac{2}{3} \text{ hour} \times 60 \text{ minutes/hour} = \frac{120}{3} \text{ minutes} = 40 \text{ minutes}. \)
So, the time taken to fill the remaining tank after 9 a.m. is 8 hours and 40 minutes.
Determining the Final Time When the Tank is Full
The time calculated (8 hours and 40 minutes) starts from 9 a.m., when all pipes began working together. To find the final time the tank is full, add this duration to 9 a.m.
Starting time: 9:00 a.m.
Add 8 hours: 9:00 a.m. + 8 hours = 5:00 p.m.
Add 40 minutes: 5:00 p.m. + 40 minutes = 5:40 p.m.
Therefore, the tank will be full at 5:40 p.m.
| Pipe | Time to fill (hours) | Rate (tank/hour) |
|---|---|---|
| A | 20 | \( \frac{1}{20} \) |
| B | 30 | \( \frac{1}{30} \) |
| C | 60 | \( \frac{1}{60} \) |
| Time Interval | Working Pipes | Combined Rate (tank/hour) | Duration (hours) | Work Done |
|---|---|---|---|---|
| 7 a.m. - 8 a.m. | A | \( \frac{1}{20} \) | 1 | \( 1 \times \frac{1}{20} = \frac{1}{20} \) |
| 8 a.m. - 9 a.m. | A + B | \( \frac{1}{20} + \frac{1}{30} = \frac{1}{12} \) | 1 | \( 1 \times \frac{1}{12} = \frac{1}{12} \) |
| After 9 a.m. | A + B + C | \( \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{1}{10} \) | \( T \) | \( T \times \frac{1}{10} \) |
Revision Table: Key Calculations
| Calculation Step | Value |
|---|---|
| Work done by 9 a.m. | \( \frac{2}{15} \) |
| Remaining work after 9 a.m. | \( \frac{13}{15} \) |
| Combined rate (A+B+C) | \( \frac{1}{10} \) tank/hour |
| Time to fill remaining tank | \( \frac{26}{3} \) hours (8 hours 40 minutes) |
| Final time | 9:00 a.m. + 8 hours 40 minutes = 5:40 p.m. |
Additional Information on Time and Work Problems
Time and work problems often involve calculating the rate at which a task is completed. The basic principle is that if someone (or something like a pipe) can complete a task in \(N\) units of time, their rate of work is \( \frac{1}{N} \) of the task per unit of time. When multiple entities work together, their individual rates are usually added to find the combined rate, assuming they work efficiently. If entities work against each other (like a leak in a tank), their rates are subtracted. Problems involving different start times require careful calculation of the work done in distinct time intervals before the final combined work phase begins. Always ensure time units are consistent (e.g., all in hours or all in minutes) throughout the calculation.
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