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
Understanding the Pipes and Cistern Problem
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.
Calculating Individual Pipe Rates
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 1 fills the cistern in 2 hours. Its filling rate is \( \frac{1}{2} \) cistern per hour.
- Pipe 2 fills the cistern in 3 hours. Its filling rate is \( \frac{1}{3} \) cistern per hour.
- Pipe 3 drains the cistern in 6 hours. Its draining rate is \( \frac{1}{6} \) cistern per hour.
| 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 | \( \frac{1}{6} \) |
Determining the Combined Rate
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.
Calculating Remaining Capacity to Fill
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.
Calculating Time to Fill the Remaining Capacity
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.
Converting Time to Hours and Minutes
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.
Revision Table: Key Calculations
| 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 |
Additional Information on Pipe and Cistern Problems
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).
- Positive Work: Pipes that fill the cistern do positive work. Their rates are added.
- Negative Work: Pipes that drain the cistern do negative work. Their rates are subtracted from the filling rates.
- Efficiency/Rate: A faster pipe (one that takes less time) has a higher filling rate.
- Total Work: The total work is usually considered as '1 unit' (representing a completely full cistern).
- Partial Work: If the cistern is partially filled or needs to be partially filled, calculate the remaining fraction of work.
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.
Similar Questions
Two pipes A and B can fill an empty cistern in 32 and 48 hours, respectively. Pipe C can drain the entire cistern in 64 hours when no other pipe is in operation. Initially, when the cistern was empty Pipe A and Pipe C were turned on. After a few hours, Pipe A was turned off and Pipe B was turned on instantly. In all it took 112 hours to fill the cistern. For how many hours was Pipe B turned on?
Pipes A, B and C are attached to an empty cistern. While the first two can fill the cistern in 4 and 10 hours, respectively, the third can drain the cistern, when filled, in 6 hours. If all the three pipes are opened simultaneously when the cistern is three-fifth full, how many hours will be needed to fill the cistern?
One pipe can fill an empty cistern in 4 hours while another can drain the cistern when full in 10 hours. Both the pipes were turned on when the cistern was half-empty. How long will it take the cistern to be full?
A pipe, working at full speed, can fill an empty cistern in 1 hour. However, during the first hour it worked at one-twelfth of its capacity, during the second hour at one-ninth of its capacity, during the third hour at one-sixth of its usual capacity, during the fourth hour at one- fourth of its usual capacity and during the fifth hour it was only one-third as efficient as it was supposed to be. A second pipe also displayed similar performance, but if it worked at full speed would have filled the empty cistern in 2 hours. Together with a drain pipe that drained water out of the tank at a constant rate, the empty cistern could be filled in 5 hours, all the three pipes working concurrently. How many hours will it take the drain pipe to empty the filled cistern if no other pipe was functioning during the time?
One of the two inlet pipes works twice as efficiently as the other. The two, working alongside a drain pipe that can empty a cistern all by itself in 8 hours, can fill the empty cistern in 8 hours. How many hours will the less efficient inlet pipe take to fill the empty cistern by itself?
A tanker can fill a cistern in 12 hours. After half the cistern is filled, 2 more similar tankers are opened. What is the time taken to fill the remaining half of the tank?
P, Q and R are channels which discharge solutions A, B, C respectively in a tank. When the tank is empty and all the three channels are opened what is the proportion of the solution C in the tank after 3 minutes, if the channels P, Q and R can fill the tank from empty to full in 30 minutes and 20 minutes and 10 minutes respectively when they are opened one at a time.
A sump is filled by three tankers with uniform flow. The first two tankers operating simultaneously fill the sump in the same time during which the sump is filled by the third tanker alone. The second tanker fills the sump 5 hours faster than the first tanker and 4 hours slower than the third tanker. The time required by the first tanker is:
Pipe A can fill a tank in 6 hours. Pipe B can empty it in 15 hours. If both the pipes are opened together, then the tank will be filled in how many hours?
- 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?
Important Questions from Pipe and Cistern
Two pipes A and B can independently fill a tank completely in 20 and 30 minutes respectively. If both the pipes are opened simultaneously, how much time will they take to fill the tank completely?
The compound interest on Rs. 64,000 for 3 years, compounded annually at 7.5% p.a. is
A water tank can be emptied in 40 minutes by a pipe of d 'diameter, so how long will it take for a 2d diameter pipe to be emptied?
A pipe can fill a tank in 4 hours, while a leak which is at one-fourth of the height of the tank from bottom can empty upto that part in 2 hours. If both are operated simultaneously and initially the tank is full, then when it will be one-fourth full?
Two pipes X and Y can fill an empty tank in 16 hours and 20 hours respectively. Pipe Z alone can empty the completely filled tank in 25 hours. Firstly both pipes X and Y are opened and after 6 hours pipe Z is also opened. What will be the total time (in hours) taken to completely fill the tank?
