A pipe can fill a tank in 30 hours. Due to a leakage at the bottom, it is filled in 50 hours. How much time will the leakage take to empty the completely filled tank?
75 hours
Understanding Tank Filling and Emptying Problems
This problem involves calculating the time taken by a leak to empty a tank, considering the effect it has on the filling time of a pipe. These types of questions are common in time and work or pipe and cistern problems in quantitative aptitude.
Analyzing the Given Information
We are given the following information:
- Time taken by the pipe alone to fill the tank = 30 hours.
- Time taken by the pipe and the leak together to fill the tank = 50 hours.
We need to find the time taken by the leak alone to empty the completely filled tank.
Calculating the Rates of Work
In such problems, we consider the amount of work done (filling or emptying) in one unit of time, usually one hour. This is called the rate of work.
- Rate of filling by the pipe = \( \frac{1}{\text{Time taken by pipe alone}} \)
- Rate of filling by the pipe = \( \frac{1}{30} \) of the tank per hour.
When the leak is present, the tank fills slower. This means the leak is doing negative work (emptying) simultaneously. The net rate of filling when both the pipe and the leak are operational is:
- Net rate of filling = \( \frac{1}{\text{Time taken by pipe and leak together}} \)
- Net rate of filling = \( \frac{1}{50} \) of the tank per hour.
Setting up the Equation
The net rate of filling is the rate of the pipe minus the rate of the leak (since the leak empties). Let 'r' be the rate at which the leak empties the tank (amount emptied per hour).
Net Rate = Rate of Pipe - Rate of Leak
\( \frac{1}{50} = \frac{1}{30} - r \)
Finding the Leakage Rate
Now, we need to solve this equation for 'r'.
\( r = \frac{1}{30} - \frac{1}{50} \)
To subtract these fractions, we find a common denominator, which is the least common multiple (LCM) of 30 and 50. The LCM of 30 and 50 is 150.
\( r = \frac{1 \times 5}{30 \times 5} - \frac{1 \times 3}{50 \times 3} \)
\( r = \frac{5}{150} - \frac{3}{150} \)
\( r = \frac{5 - 3}{150} \)
\( r = \frac{2}{150} \)
\( r = \frac{1}{75} \)
So, the rate of the leak is \( \frac{1}{75} \) of the tank emptied per hour.
Determining the Time to Empty the Tank
If the leak empties \( \frac{1}{75} \) of the tank in one hour, then the time taken to empty the entire tank (which is 1 whole tank) is the reciprocal of the rate.
Time taken by leak to empty = \( \frac{1}{\text{Rate of leak}} \)
Time taken by leak to empty = \( \frac{1}{1/75} \)
Time taken by leak to empty = \( 1 \times 75 \)
Time taken by leak to empty = 75 hours.
Therefore, the leakage will take 75 hours to empty the completely filled tank.
| Action | Time (hours) | Rate (tank/hour) |
|---|---|---|
| Pipe Filling Alone | 30 | \( \frac{1}{30} \) |
| Pipe + Leak Filling (Net) | 50 | \( \frac{1}{50} \) |
| Leak Emptying Alone | ? | \( r = \frac{1}{75} \) |
This table summarizes the rates involved in the tank filling and leakage problem.
Revision Table: Tank and Leak Concepts
| Concept | Explanation | Formula |
|---|---|---|
| Work Rate | Amount of work done per unit time (e.g., tank filled per hour). | Rate = 1 / Time |
| Filling Pipe Rate | Positive rate, adds liquid to the tank. | \( R_{fill} = \frac{1}{T_{fill}} \) |
| Leak Rate | Negative rate, removes liquid from the tank. | \( R_{leak} = \frac{1}{T_{leak}} \) |
| Net Rate (Fill Pipe + Leak) | Combined effect; rate of filling minus rate of emptying. | \( R_{net} = R_{fill} - R_{leak} \) |
| Time for Leak Alone | Time taken by the leak to empty the full tank. | \( T_{leak} = \frac{1}{R_{leak}} \) |
Additional Information: Solving Time and Work Problems
Problems involving pipes, cisterns, and leaks are applications of the time and work concept. The fundamental idea is that the total work done is equal to the rate of work multiplied by the time taken.
- If a task is completed in 'T' hours, the rate of work is 1/T per hour.
- If multiple entities work together, their rates are added if they do the same type of work (e.g., two pipes filling) or subtracted if they do opposite work (e.g., a pipe filling and a leak emptying).
- Total work is often considered as '1' unit (representing the full tank or the complete task).
- Understanding the rates is key to solving these problems efficiently.
Always pay attention to whether the work is positive (filling) or negative (emptying) when combining rates.
Similar Questions
An inlet pipe can fill an empty tank in 140 hours while an outlet pipe drains a completely-filled tank in 63 hours. If 8 inlet pipes and y outlet pipes are opened simultaneously, when the tank is empty, then the tank gets completely filled in 105 hours. Find the value of y.
There are two inlet pipes A and B connected to a tank. A and B can fill the tank in 32 h and 28 h, respectively. If both the pipes are opened alternately for 1 h, starting with A, then in how much time (in hours, to nearest integer) will the tank be filled?
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Two pipes S1 and S2 alone can fill an empty tank in 15 hours and 20 hours respectively. Pipe S3 alone can empty that completely filled tank in 40 hours. Firstly both pipes S1 and S2 are opened and after 2 hour pipe S3 is also opened. In how much time tank will be completely filled after S3 is opened?
Pipe A and pipe B running together can fill a cistern in 6 minutes. If B takes 5 minutes more than A to fill it, then the time in which A and B will fill that cistern separately will be, respectively, __________ .
There are 3 taps A, B, and C in a tank. These can fill the tank in 10 hours, 20 hours and 25 hours, respectively. At first, all three taps are opened simultaneously. After 2 hours, tap C is closed and A and B keep running. After 4 hours from the beginning, tap B is also closed. The remaining tank is filled by tap A alone. Find the percentage of work done by tap A itself.
There are two water taps in a tank which can fill the empty tank in 12 hours and 18 hours respectively. It is seen that there is a leakage point at the bottom of the tank which can empty the completely filled tank in 36 hours. If both the water taps are opened at the same time to fill the empty tank and leakage point was repaired after 1 hour, then in how much time the empty tank will be completely filled?
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?
