Introduction: Pipes & Cisterns
- Pipes and Cisterns problems are almost the same as those of Time and Work problems.
- Thus, if a pipe fills a tank in 6 hrs, then the pipe fills 1/6th of the tank in 1 hour.
- The only difference with Pipes and Cisterns problems is that there are outlets as well as inlets.
- Thus, there are agents (the outlets) which perform negative work too.
- The rest of the process is almost similar.
Inlet
- A pipe connected with a tank (or a cistern or a reservoir) is called an inlet, if it fills the tank. The work done by this is taken as Positive work.
Outlet
- A pipe connected with a tank is called an outlet if it empties the tank and the work done by this is taken as the Negative work.
Basic Concept
- The basic concept of Time and Work is similar to that across all arithmetic topics, i.e. the concept of proportionality.
- Efficiency is inversely proportional to the Time taken when the amount of work done is constant.
- This can be used to compare efficiencies and time taken across different groups in Time-Speed-Distance, efficiency is replaced by Speed, i.e. Speed is inversely proportional to Time when the Distance is constant.
- Pipes and Cisterns are just an application of Time and Work. Concept wise, it is one and the same.
- In the above proportionality, Efficiency is replaced by Rate of filling.
The equation, in this case, becomes:
Question for Solved Examples: Time & Work
Try yourself:Three pipe P, Q and R can fill a tank In 12 minutes, 18 minutes and 24 minutes respectively. The pipe R is closed 12 minutes before the tank Is filled. In what time the tank is full?
Explanation
Let T is the time taken by the pipes to fill the tank
Let capacity of tank = L C M o f ( 12 , 18 , 24 ) = 72
∴ P in 1 min can fill 6 units
Q in 1 min can fill 4 units
R in 1 min can fill 3 units
∴ ( 10 t + 3 ( t − 12 ) = 72
13 t = 108
⇒ t =
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Introduction: Negative Work
- Negative work increases the time in which work is to be completed. This application can be extended to cases involving Pipes and cisterns.
- Suppose there are two pipes in a Cistern. Pipe A is used to fill the Cistern and Pipe B is used to empty the Cistern. Here we say that Pipe B and Pipe A are working against each other. When a leak is developed in the Cistern, the leak forms the component of negative work, which slows down the completion of the task (in this case, the filling of the Cistern).
Example 1. A Cistern has three pipes A, B, and C. Pipe A can fill a Cistern in 10 hrs, Pipe B can fill a Cistern in 5 hrs while Pipe C can empty the Cistern in 20 hrs. If they are switched on at the same time; in how many hours will the Cistern be filled?
Solution:
- In one hour, Pipe A can fill 100/10 = 10% of the Cistern.
- In one hour, Pipe B can fill 100/5 = 20 % of the Cistern.
- In one hour, Pipe C can empty 100/20 = 5 % of the Cistern.
- If all three are working together, (10 + 20 – 5) = 25% of the Cistern will get filled in one hour, so it will take 4 hrs for the Cistern to fill.
Example 2. A tank can be filled by tap A in 6 hrs and by tap B in 3 hrs. But when they are open simultaneously to fill an empty tank, they take 3 hrs more than their normal Time. A hole is later discovered as the reason for the delay. Find the Time taken by the hole to empty the tank if it is completely filled.
Solution:
- Tap A takes 6 hrs to fill, so in one hour it will fill 16.67 % of the tank.
- Tap B takes 3 hrs to fill, so in one hour it will fill 33.33% of the tank.
- Together they will fill 16.67% + 33.33% = 50 % of the tank.
- They should take 100/50= 2 hrs to fill the tank. But they take 3 hrs more, because of the hole; they totally take 5 hrs to fill, i.e. they fill 20% in an hour.
- This is possible if the hole empties 50-20 = 30% of the tank in an hour.
- So, to completely empty the full tank, the hole will take 100/30 hrs = 3.33 hrs = 3 hrs and 20 mins.
Example 3. Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
Solution:
- Part filled by A alone in 1 hour = 1/36
- Part filled by B alone in 1 hour = 1/45
∴ Part filled by (A + B) in 1 hour =
- Hence, both the pipes together will fill the tank in 20 hours.
Question for Solved Examples: Time & Work
Try yourself:Pipe A can fill a tank in ‘an’ hours. On account of a leak at the bottom of the tank, it takes thrice as long to fill the tank. How long will the leak at the bottom of the tank take to empty a full tank, when pipe A is kept closed?
Explanation
Method 1: Using variables
The pipe can fill (1/a)th of the tank in an hour. Because of the leak, it can only fill 1/3a of the tank per hour. Let X be the time in which the leak can completely empty the tank, hence
1/x=1/a-1/3a = x =(3a/2)hrs.
Answer: Option (a)
Method 2: Using Numbers (Shortcut)
Assume a value for “a”- say 10 hours. Because of the leak, it will take 30 hours. Now, this means that in these 30 hours, the filling will occur at a rate of 3.33% and the leaking will slow down the process by 6.66% every hour Thus, the Time taken to empty a full tank= 100/6.66= 15 hours.
Answer: Only option (a) satisfies this value.
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Question for Solved Examples: Time & Work
Try yourself:Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution R in the liquid in the tank after 3 minutes?
Explanation
Explanation:
Part filled by (A + B + C) in 3 minutes = 3(1/30 + 1/20 + 1/10) = 11/20
Part filled by C in 3 minutes = 3/10
Required ratio = 3/10 * 20/11 = 6/11
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Question for Solved Examples: Time & Work
Try yourself:Pipe A can fill a tank in 36 hours and pipe B can empty it in 45 hours. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
Solution.
Explanation
Total Volume of the tank = 180 units (LCM of 36 and 45)
1 hour work of A = 5 units
1 hour work of B = - 4 units
1 hour work of A + B = 1 units
So, Total time taken = 180 hours
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