Pipes and Cistern Examples 1

In this class, We discuss Pipes and Cistern Examples 1.

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Example 1:

The reservoir fills in 12 hours if two pipes function simultaneously.

One pipe fills the reservoir 10 hours faster than the other.

How many hours does it take the second pipe to fill the reservoir?

Solution:

Assume the First pipe takes x hours.

The second pipe will take x+10 hours.

From the formulae (1/x + 1/(x+10)) = 1/12

(x+10+x)/x(x+10) = 1/12

x^2 – 14x -120 = 0

x=20.

The second pipe will take 20 + 10 hours.

= 30 hours.

Example 2:

Two pipes can fill a tank in 14 and 16 hours, respectively.

The pipes opened simultaneously, and it was found the leakage in the bottom took 32 minutes more time to fill the tank.

When the tank is full, what time will the leakage empty it?

Solution:

Two pipes work simultaneously. We have the formulae (A + B) ‘s one hour = (1/14 + 1/15) = 15/112.

Time taken to fill is 112/15

= 7 hours 28 minutes

Due to the leak, 32 minutes extra time to fill the tank.

Total time = 7.28 + 32 minutes. = 8 hours.

Leak flow per hour = 15/112 – 1/8 = 1/112

The leak will empty in 112 hours.