Two pipes, when working one at a time can fill a cistern in 2 hours and 3 hours, respectively while a third pipe can drain the cistern empty in 6 hours. All the three pipes were opened together when the cistern was 1/6 full. How long will it take for the cistern to be completely full?
1 hour 15 minutes
This problem involves pipes that fill a cistern and a pipe that drains it. We need to find out how long it takes to fill the cistern completely when it is already partially filled and all pipes are working together. These types of problems are often solved by calculating the work rate of each pipe.
The rate at which a pipe fills or drains a cistern is usually expressed as the fraction of the cistern filled or drained in one hour. If a pipe fills a cistern in 'T' hours, its filling rate is \( \frac{1}{T} \) cistern per hour. If a pipe drains a cistern in 'T' hours, its draining rate is \( \frac{1}{T} \) cistern per hour (but its contribution is negative).
| Pipe | Type | Time (hours) | Rate (cistern/hour) |
|---|---|---|---|
| Pipe 1 | Filling | 2 | \( \frac{1}{2} \) |
| Pipe 2 | Filling | 3 | \( \frac{1}{3} \) |
| Pipe 3 | Draining | 6 | \( \frac{1}{6} \) |
When all pipes work together, their rates are combined. Filling rates are positive, and draining rates are negative. The combined rate is the sum of the individual rates:
Combined Rate = (Rate of Pipe 1) + (Rate of Pipe 2) - (Rate of Pipe 3)
Combined Rate = \( \frac{1}{2} + \frac{1}{3} - \frac{1}{6} \)
To add and subtract these fractions, we need a common denominator, which is 6.
\( \frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6} \)
\( \frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6} \)
\( \frac{1}{6} = \frac{1 \times 1}{6 \times 1} = \frac{1}{6} \)
Combined Rate = \( \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{3+2-1}{6} = \frac{4}{6} = \frac{2}{3} \) cistern per hour.
The combined rate is \( \frac{2}{3} \) cistern per hour, meaning that with all three pipes open, \( \frac{2}{3} \) of the cistern is filled every hour.
The cistern was initially \( \frac{1}{6} \) full. To be completely full (1 whole cistern), the remaining capacity that needs to be filled is:
Remaining Capacity = Total Capacity - Initially Filled
Remaining Capacity = \( 1 - \frac{1}{6} \)
Remaining Capacity = \( \frac{6}{6} - \frac{1}{6} = \frac{6-1}{6} = \frac{5}{6} \) cistern.
So, \( \frac{5}{6} \) of the cistern still needs to be filled.
The time required to fill the remaining part of the cistern is found by dividing the remaining capacity by the combined filling rate:
Time = \( \frac{\text{Remaining Capacity}}{\text{Combined Rate}} \)
Time = \( \frac{5/6 \text{ cistern}}{2/3 \text{ cistern/hour}} \)
To divide by a fraction, we multiply by its reciprocal:
Time = \( \frac{5}{6} \times \frac{3}{2} \) hours
Time = \( \frac{5 \times 3}{6 \times 2} = \frac{15}{12} \) hours
Simplify the fraction \( \frac{15}{12} \) by dividing both numerator and denominator by 3:
Time = \( \frac{15 \div 3}{12 \div 3} = \frac{5}{4} \) hours.
The time is \( \frac{5}{4} \) hours. This can be written as a mixed number:
\( \frac{5}{4} = 1 \frac{1}{4} \) hours.
This is 1 full hour plus \( \frac{1}{4} \) of an hour. To convert \( \frac{1}{4} \) hour to minutes, multiply by 60:
\( \frac{1}{4} \) hour \( = \frac{1}{4} \times 60 \) minutes \( = \frac{60}{4} \) minutes \( = 15 \) minutes.
Therefore, the total time required to fill the remaining \( \frac{5}{6} \) of the cistern is 1 hour and 15 minutes.
| Calculation Step | Formula/Operation | Result |
|---|---|---|
| Pipe 1 Rate | 1 / Time | \( \frac{1}{2} \) cistern/hour |
| Pipe 2 Rate | 1 / Time | \( \frac{1}{3} \) cistern/hour |
| Pipe 3 Rate | 1 / Time | \( \frac{1}{6} \) cistern/hour |
| Combined Rate | Rate1 + Rate2 - Rate3 | \( \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{2}{3} \) cistern/hour |
| Remaining Capacity | Total - Initial Fill | \( 1 - \frac{1}{6} = \frac{5}{6} \) cistern |
| Time to Fill Remaining | Remaining Capacity / Combined Rate | \( \frac{5/6}{2/3} = \frac{5}{4} \) hours |
| Convert Hours to Minutes | Fractional Hours x 60 | \( \frac{1}{4} \times 60 = 15 \) minutes |
| Final Time | Full Hours + Minutes | 1 hour 15 minutes |
Pipe and cistern problems are a common type of quantitative aptitude question that falls under the broader category of 'Time and Work'. The core idea is to calculate the fraction of work done (or cistern filled/drained) per unit of time (usually an hour or a minute).
These problems often require comfort with fractions and converting between hours and minutes. Always ensure you are clear whether a pipe is filling or draining and whether you are calculating the time for the entire cistern or just a portion of it.
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