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Pipes and Cisterns Concept
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This lesson discusses the concept of pipes n cisterns and what are the different kind of questions asked from this topic

## Jaspreet Kaur is teaching live on Unacademy Plus

Jaspreet Kaur
Computer Science Engineer , Web Developer , Programmer , Cleared GATE Examination twice

U
Sir pls kindly make a course on current affairs from Jan to Nov
1. Concept of Pipes and Cisterns

2. HELLO! I am Jaspreet Kaur B.tech in Computer Science and Engineering ,Pantnagar University. Interests : Computer Programming, Engineering, Web Development Problem solving b https://unacademy.com/user/ jaspreet.kaur309-6837

3. Example : Two pipes A and B can fill a cistern in 20 and 30 minutes respectively, and a third pipe C can empty it in 40 minutes. How much time will it take to fill the cistern if all the three are opened at the same time? Solution : Problem LCM of 20, 30, and 40 = 120, Let us assume that the capacity of the cistern is 120 liters. Therefore, Rate of pipe A = 120/20 = 6 litres/min

4. Similarly, Rate of pipe B = 120/30-4 liters/min And rate of pipe C 120/40 3 liters/min All three pipes are opened simultaneously, then in a min, the amount of water in the cistern in a min will be 6+4-3 7 liters. As pipe C is an outlet pipe, we are taking its work to be negative Hence, the time taken to fill the cistern 120/7 min.

5. Shortcut 1 The time required by leak alone to empty the cistern If an inlet pipe takes X min to fill a cistern, but due to leak, it takes Y extra min to fill the cistern, then the amount of time in which the leak can empty the full cistern with the inlet pipe, not in operation will be (X2 XY)/ Y min

6. Example: If a cistern generally takes 20 min to be filled by a pipe, but due to leak, it takes 10 extra min to be filled, then the amount of time in which the leak can empty the full cistern is: Solution: Example Normal Method :Inlet pipe alone takes 20 min. (Inlet pipe Leak) take 20+10 30 min. LCM of 20 and 30 60. Let the capacity of the cistern be 60 liters. Rate per minute of inlet pipe60/20 3 liters/min (1)

7. Rate per minute of inlet +leak 60/30 2 iters/min ..(2) solving (1) and (2) we get the rate per minute of leak =-1 liters/min Therefore, when the inlet is non-operational, the leak will take 60/1 = 60 minutes or one hour to empty the full cistern.

8. shortcut Trick and Form20 and Y-10. With Trick On comparing, we have X-20 and Y=10. Therefore, the leak will take ( 202 + 20 X 10 )/ 10 = 60 min

9. Shortcut 2 Finding the capacity of the cistern A cistern has a leak which can empty it in X hours. A pipe which admits Y liters of water per hour into the cistern is turned on, and now the cistern is emptied in Z hours. Then the capacity of the cistern is liters X*Y*Z/ (Z - X)

10. Example: A cistern has a leak which can empty it in 4 hours. A pipe which admits 20 liters of water per hour into the cistern is turned on, and now the cistern is emptied in 6 hours. What is the capacity of the cistern? Solution Example Shortcut Trick and Formula: By comparing with the shortcut formula, X-4, Y-20, and Z-6. Therefore, the capacity of the cistern 4 20 6/6-4) 240 Or the capacity of the cistern = 240 liters. 10

11. Example : Two pipes A and B can fill a cistern in 12 and 16 min respectively. Both pipes are opened together but 4 min before the cistern is full, one pipe A is closed. Find the time required to fill the cistern. Solution Example Shortcut Trick and Formula: Let it takes x min by the pipes A and B to fill the cistern. Therefore, the duration for which pipe was operating = (x-4) min. And pipe B was operating for x min. 12

12. Shortcut 4 The rate of flow of the pipe is proportional to the diameter. 14