Aptitude formula - Pipes and Cisterns - ObjectiveBooks

Aptitude formula - Pipes and Cisterns

Fastest way to solve Aptitude problems, shortcuts, tricks and important formulas - Pipes and Cisterns:

Inlet: A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.
Outlet: A pipe connected with a tank or a cistern or a reservoir, emptying it, is known as an outlet.

1. If a pipe can fill a tank in Î± hours, then: part filled in 1 hour = 1/Î±

2. If a pipe can empty a full tank in Î² hours, then: part emptied in 1 hour = 1/Î²

3. If a pipe can fill a tank in Î± hours and another pipe can empty the full tank in Î² hours (where Î²> Î±), then on opening both the pipes,
The net part filled in 1 hour = (1/Î±) - (1/Î²)

4. If a pipe can fill a tank in Î± hours and another pipe can empty the full tank in Î² hours (where Î± > Î²), then on opening both the pipes,
The net part emptied in 1 hour = (1/Î²) - (1/Î±)

Also, these shortcut formulas can be used:

1. If a pipe can fill a tank in Î± hours and another pipe can empty the full tank in Î² hours (where Î²> Î±), then time taken to fill the tank, when both the pipe are opened, Î±Î²/(Î² - Î±)

2. If a pipe can fill a tank in Î± hours and another pipe can fill the same tank in Î² hours, then the net part the time taken to fill, when both the pipes are open
Time taken to fill the tank = Î±Î²/(Î± + Î²)

3. If a pipe fills a tank in Î± hour and another fills the same tank in Î² hours, but a third one empties the full tank in É£ hours, and all of them are opened together, then
Time taken to fill the tank = Î±Î²É£/(Î²É£ +Î±É£ - Î±Î²)

4. A pipe can fill a tank in Î± hours. Due to a leak in the bottom, it is filled in Î² hours. The time taken by the leak to empty the tank is, Î±Î²/(Î² - Î±)

Example: 01
Two pipes A and B can fill the tank in 30 hrs and 45 hrs respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

Solution:
By 1st method;
A fills the tank in 1 hr = 1/30 parts
B fills the tank in 1 hr = 1/45 parts
A and B together fills the tank in 1 hr = 1/30 + 1/45 = 1/18 parts
So, time required to fill the tank is 18 hrs.

By 2nd method; Time taken = Î±Î²/(Î± + Î²) = (30 × 45)/(30 + 45) = 18 hrs.

Example: 02
Pipe A can fill a tank in 25 hrs while B alone can fill it in 30 hrs and C can empty the full tank in 45 hrs. If all the pipes are opened together, how much time will be needed to make the tank full?

Solution:
The tank will be full in = (25 × 30 × 45)/[(30 × 45) + (25 × 45) - (25 × 30)] = 19.56 hours.

Example: 03
Two pipes A and B would fill a cistern in 24 hrs and 32 hrs respectively. If both the pipes are opened together; find when the first pipe must be turned off so that the cistern may be just filled in 16 hrs.

Solution:
B fills the tank in 1 hr = 1/32 parts
B fills the tank in 16 hrs = 16/32 parts = 1/2 part.
Given,
A fill the full tank in 24 hrs
Therefore, A fills the 1/2 part in just 12 hrs.
So, the first pipe A should work for 12 hrs.

Alternate Method: The first should work for = [1 - (16/32)] × 24 = 12 hrs.

Pipes and Cisterns:
Formula:                Pipes and Cisterns Formulas
Solved Examples:  Solved Examples: Set 01
Practice Test:         Practice Test: 01
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