Two valves A and B can fill a sump in \(37\frac{1}{2}\) minutes and 45 minutes respectively. Both valves are opened. The sump will be filled in just 30 minutes, if valve B is turned off after?
9 min
Solving the Sump Filling Problem with Valves A and B
This problem involves calculating the time a valve should be active when multiple valves are working together to fill a sump. We need to determine when valve B, which fills the sump in 45 minutes, must be turned off so that valve A (filling in \(37\frac{1}{2}\) minutes) and valve B together fill the sump in exactly 30 minutes.
Understanding the Concept of Work Rate
In problems like this, the rate of work (or filling) is the amount of work done (part of the sump filled) per unit of time. If a valve can fill a sump in \(T\) minutes, its rate is \(\frac{1}{T}\) of the sump per minute.
Calculating Individual Valve Rates
- Valve A fills the sump in \(37\frac{1}{2}\) minutes.
- First, convert the mixed number to an improper fraction: \(37\frac{1}{2} = 37 + \frac{1}{2} = \frac{74}{2} + \frac{1}{2} = \frac{75}{2}\) minutes.
- Rate of Valve A = \(\frac{1}{\text{Time taken by A}} = \frac{1}{\frac{75}{2}} = \frac{2}{75}\) of the sump per minute.
- Valve B fills the sump in 45 minutes.
- Rate of Valve B = \(\frac{1}{\text{Time taken by B}} = \frac{1}{45}\) of the sump per minute.
Setting up the Equation for Total Work
Both valves start together. Let's say valve B is turned off after \(t\) minutes. The total time taken to fill the sump is 30 minutes.
- Valve A works for the entire duration, which is 30 minutes.
- Valve B works for the first \(t\) minutes.
The total work done (filling the entire sump) is equal to the sum of the work done by valve A and valve B.
Work done by A = (Rate of A) \(\times\) (Time A worked)
Work done by B = (Rate of B) \(\times\) (Time B worked)
Total Work = Work done by A + Work done by B
Since the entire sump is filled, Total Work = 1.
\[ \left(\frac{2}{75} \times 30\right) + \left(\frac{1}{45} \times t\right) = 1 \]Solving for \(t\)
Now, let's solve the equation to find the value of \(t\):
First, simplify the first term:
\[ \frac{2}{75} \times 30 = \frac{60}{75} \]We can simplify \(\frac{60}{75}\) by dividing both the numerator and denominator by their greatest common divisor, which is 15:
\[ \frac{60 \div 15}{75 \div 15} = \frac{4}{5} \]So the equation becomes:
\[ \frac{4}{5} + \frac{t}{45} = 1 \]To isolate the term with \(t\), subtract \(\frac{4}{5}\) from both sides of the equation:
\[ \frac{t}{45} = 1 - \frac{4}{5} \]To subtract the fractions on the right side, find a common denominator, which is 5:
\[ 1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{5-4}{5} = \frac{1}{5} \]So the equation is now:
\[ \frac{t}{45} = \frac{1}{5} \]To find \(t\), multiply both sides of the equation by 45:
\[ t = \frac{1}{5} \times 45 \] \[ t = \frac{45}{5} \] \[ t = 9 \]The value of \(t\) is 9 minutes.
This means that valve B was working for 9 minutes before being turned off.
Conclusion
For the sump to be filled in exactly 30 minutes, valve B must be turned off after 9 minutes.
| Valve | Time to Fill (minutes) | Rate (part/minute) |
|---|---|---|
| A | \(37\frac{1}{2} = \frac{75}{2}\) | \(\frac{2}{75}\) |
| B | 45 | \(\frac{1}{45}\) |
Revision Table: Sump Filling Calculation
| Step | Description | Calculation/Equation |
|---|---|---|
| 1 | Rate of Valve A | \( \frac{1}{37.5} = \frac{2}{75} \) |
| 2 | Rate of Valve B | \( \frac{1}{45} \) |
| 3 | Equation for total work (let \(t\) be time B is on) | \( \left(\frac{2}{75} \times 30\right) + \left(\frac{1}{45} \times t\right) = 1 \) |
| 4 | Simplify and Solve for \(t\) | \( \frac{60}{75} + \frac{t}{45} = 1 \Rightarrow \frac{4}{5} + \frac{t}{45} = 1 \Rightarrow \frac{t}{45} = \frac{1}{5} \Rightarrow t = 9 \) |
Additional Information: Pipes and Cisterns Problems
Problems involving pipes and cisterns are similar to time and work problems. The concepts of rates and total work done apply here. Here are some key points:
- Filling Pipes: These pipes do positive work (filling the tank/sump). Their rates are added when working together.
- Emptying Pipes: These pipes do negative work (emptying the tank/sump). Their rates are subtracted from the filling rates when working simultaneously.
- Combined Rate: When multiple pipes work together, their individual rates are combined (added for filling, subtracted for emptying) to find the combined rate.
- Total Time: If the combined rate is \(R\) parts per unit of time, the total time taken to fill or empty the entire tank (1 part) is \(\frac{1}{R}\) units of time.
- Partial Work: If a pipe works for a specific duration, the part of the tank filled or emptied is (Rate \(\times\) Time). The sum or difference of partial work done by each pipe gives the total work done.
Understanding these basic principles helps in setting up the correct equations for various scenarios involving pipes working together or turning off at different times.
Similar Questions
Three pipes A, B and C can fill a tank in 12 hours, 18 hours and 24 hours, respectively. A leak at the bottom can empty the full tank in 36 hours. If all the pipes are opened together, in how many hours will the tank be filled? (Round off your answer to two decimal places.)
Pipes A and B can fill an entire tank in 8 hours and 12 hours, respectively. The water tank is one-fourth full. If both the pipes are opened together, then how long will it take to fill the remaining part of the tank?
A pump can fill a tank in 3 hours. Due to a leak in the tank, it takes 4.5 hours to fill the tank. In how much time can the leak empty the full tank if no other entry or exit routes are open?
Two pipes, when working one at a time, can fill a cistern in 3 hours and 4 hours, respectively while a third pipe can drain the cistern empty in 8 hours. All the three pipes were opened together when the cistern was 1/12 full. How long did it take for the cistern to be completely full?
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?
Pipes A and C can fill an empty cistern in 16 and 24 hours respectively while Pipe B can drain the filled cistern in 12 hours. If the three pipes are turned on together when the cistern is empty, how many hours will it take for the cistern to be full?
Pipes A and C can fill an empty cistern in 32 and 48 hours, respectively while pipe B can drain the filled cistern in 24 hours. If the three pipes are turned on together when the cistern is empty, how many hours will it take for the cistern to be 2/3 full?
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?
- 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
A pipe can fill a cistern in 20 minutes where as the cistern when full can be emptied by a leak in 28 minutes. When both are opened, The time taken to fill the cistern is:
‘A’ pipe can empty a tank in 20 minutes. The second pipe ‘B’ has a diameter twice as that of ‘A’. If both A & B pipe are attached to the tank how much time will be required to empty the tank?
Pipes A and B can empty a full tank in 16 hours and 24 hours, respectively. Pipe C alone can fill the empty tank in 4 hours. If A, B and C are opened together, the tank will be 35% full after :
Pipes A and B can fill a tank in 36 minutes and 45 minutes, respectively. Both these pipes were opened simultaneously. After 20 minutes, a leak at the bottom of the tank was spotted which was immediately sealed. The tank was full in another 15 minutes. The leak alone can empty the full tank in:
A cistern has a leak which would empty it in 6 hours. A tap is turned on which admits 10 litres of water per minute into the cistern. When it is full it is now emptied in 10 hours. What is the capacity (in litres) of the cistern?
