Pipes A and B can fill a tank in 10 hours and 40 hours respectively. C is an outlet pipe attached to the tank. If all the three pipes are opened simultaneously, it takes 80 minutes more time than A and B together takes to fill the tank. If A and B kept open for 7 hours and closed and then C opened. How much time will C take to empty the tank :
49 hours
Solving the Pipe and Tank Filling and Emptying Problem
This problem involves understanding the rates at which pipes fill or empty a tank. We are given the filling times for two inlet pipes A and B, and information about an outlet pipe C when all three work together, as well as a scenario where A and B fill partially before C is opened to empty the tank.
Understanding Pipe Rates
The rate of a pipe is the fraction of the tank it can fill or empty in one hour. If a pipe can fill a tank in 't' hours, its filling rate is \( \frac{1}{t} \) tank per hour. If a pipe can empty a tank in 't' hours, its emptying rate is \( -\frac{1}{t} \) tank per hour (negative sign indicates emptying).
- Pipe A fills the tank in 10 hours. Rate of A \( = \frac{1}{10} \) tank/hour.
- Pipe B fills the tank in 40 hours. Rate of B \( = \frac{1}{40} \) tank/hour.
Calculating Combined Filling Time for A and B
When pipes A and B work together, their rates add up.
Combined rate of A and B \( = \) Rate of A \( + \) Rate of B
\[ \text{Combined rate of A and B} = \frac{1}{10} + \frac{1}{40} \]To add these fractions, we find a common denominator, which is 40.
\[ \frac{4}{40} + \frac{1}{40} = \frac{4+1}{40} = \frac{5}{40} = \frac{1}{8} \text{ tank/hour} \]The time taken by A and B together to fill the tank is the reciprocal of their combined rate.
Time for A and B together \( = \frac{1}{\text{Combined rate of A and B}} = \frac{1}{1/8} = 8 \) hours.
Analyzing the Case with Pipes A, B, and C Open Simultaneously
When A, B, and C are opened simultaneously, the problem states it takes 80 minutes more than the time A and B together take to fill the tank.
80 minutes converted to hours \( = \frac{80}{60} = \frac{8}{6} = \frac{4}{3} \) hours.
Time taken by A, B, and C together \( = \) Time for A and B together \( + \) 80 minutes
\[ \text{Time for A, B, and C together} = 8 \text{ hours} + \frac{4}{3} \text{ hours} = \frac{24}{3} + \frac{4}{3} = \frac{28}{3} \text{ hours} \]The combined rate of A, B, and C is the reciprocal of this time.
Combined rate of A, B, and C \( = \frac{1}{28/3} = \frac{3}{28} \) tank/hour.
The combined rate of A, B, and C is also the sum of their individual rates. Let the rate of outlet pipe C be \( -\frac{1}{c} \), where \( c \) is the time C takes to empty the full tank.
Combined rate of A, B, and C \( = \) Rate of A \( + \) Rate of B \( + \) Rate of C
\[ \frac{3}{28} = \frac{1}{10} + \frac{1}{40} + \left(-\frac{1}{c}\right) \]We already calculated \( \frac{1}{10} + \frac{1}{40} = \frac{1}{8} \).
\[ \frac{3}{28} = \frac{1}{8} - \frac{1}{c} \]Now, we solve for \( \frac{1}{c} \):
\[ \frac{1}{c} = \frac{1}{8} - \frac{3}{28} \]To subtract these fractions, find a common denominator, which is 56.
\[ \frac{1}{c} = \frac{1 \times 7}{8 \times 7} - \frac{3 \times 2}{28 \times 2} = \frac{7}{56} - \frac{6}{56} = \frac{7-6}{56} = \frac{1}{56} \]So, the rate of pipe C is \( \frac{1}{56} \) tank/hour (meaning it empties \( \frac{1}{56} \) of the tank per hour). The time taken by C to empty the full tank is \( c = 56 \) hours.
Calculating Amount Filled by A and B in 7 Hours
In the second scenario, pipes A and B are kept open for 7 hours.
Amount filled by A and B in 7 hours \( = \) Combined rate of A and B \( \times \) Time
\[ \text{Amount filled} = \frac{1}{8} \text{ tank/hour} \times 7 \text{ hours} = \frac{7}{8} \text{ of the tank} \]After 7 hours, the tank is \( \frac{7}{8} \) full.
Calculating Time for C to Empty the Filled Amount
After A and B are closed, pipe C is opened to empty the tank. Pipe C will empty the amount that was filled by A and B, which is \( \frac{7}{8} \) of the tank.
The rate at which C empties is \( \frac{1}{56} \) tank/hour.
Time taken by C to empty \( \frac{7}{8} \) of the tank \( = \frac{\text{Amount to be emptied}}{\text{Rate of C}} \)
\[ \text{Time} = \frac{7/8}{1/56} \]Dividing by a fraction is the same as multiplying by its reciprocal:
\[ \text{Time} = \frac{7}{8} \times 56 \] \[ \text{Time} = 7 \times \frac{56}{8} = 7 \times 7 = 49 \text{ hours} \]So, pipe C will take 49 hours to empty the tank after A and B have filled \( \frac{7}{8} \) of it in 7 hours.
| Pipe | Role | Time | Rate (tank/hour) |
|---|---|---|---|
| A | Inlet | 10 hours | \( \frac{1}{10} \) |
| B | Inlet | 40 hours | \( \frac{1}{40} \) |
| A & B together | Inlet | 8 hours | \( \frac{1}{8} \) |
| C | Outlet | 56 hours | \( \frac{1}{56} \) |
| A, B, & C together | Net | \( \frac{28}{3} \) hours | \( \frac{3}{28} \) |
Summary of Steps
- Calculate individual rates of filling pipes A and B.
- Calculate the combined rate and time for A and B to fill the tank.
- Use the given information about A, B, and C together to find the combined rate of all three.
- Use the combined rate and the rates of A and B to find the rate of outlet pipe C.
- Calculate the amount of tank filled by A and B working for the specified time (7 hours).
- Calculate the time taken by pipe C to empty this specific amount of water.
Revision Table: Pipe and Cistern Concepts
| Concept | Description | Formula/Relation |
|---|---|---|
| Individual Rate | Fraction of work done by one pipe in unit time. | If time is T, Rate \( = \frac{1}{T} \) |
| Combined Rate (Inlets) | Sum of individual rates of filling pipes. | Rate\( _{total} = \) Rate\( _{1} + \) Rate\( _{2} + ... \) |
| Combined Rate (Inlets & Outlets) | Sum of inlet rates minus sum of outlet rates. | Rate\( _{net} = \) (Rates of Inlets) - (Rates of Outlets) |
| Time Taken | Reciprocal of the net rate if rate is for a full tank. | Time \( = \frac{1}{\text{Net Rate}} \) (for full tank) |
| Time for Partial Work | Amount of work done divided by the rate. | Time \( = \frac{\text{Amount of Tank}}{\text{Rate}} \) |
Additional Information: Solving Pipe and Tank Problems
Pipe and tank problems are a common type of question in quantitative aptitude. They are essentially variations of time and work problems. The key is to convert the given times into rates (work per unit time) and then add or subtract rates based on whether the pipes are filling or emptying.
- Always ensure all time units are consistent (e.g., all in hours or all in minutes).
- Filling is positive work, emptying is negative work.
- If a pipe works for a partial time, the amount of work done is Rate \( \times \) Time.
- If the tank is partially filled and then emptied, calculate the amount that needs to be emptied and use the emptying pipe's rate.
- If multiple pipes work together, sum their rates (inlets add, outlets subtract) to find the net combined rate.
Understanding these basic principles helps in solving complex problems involving multiple pipes working simultaneously or in stages.
Similar Questions
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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?
