Pipes and Cisterns: Solved Questions | SSC CGL Tier 2 - Study Material, Online Tests, Previous Year PDF Download

Time, Rate, and Volume Relationship:
Time = Volume / Rate
Rate = Volume / Time
Volume = Rate × Time

Example 1: There are 15 vessels of water with the volume of 60 litters each. Together they can fill the cistern completely. Calculate how many vessels will be required to completely fill the cistern if their volume is reduced to 20 liters.
(a) 40 vessels
(b) 45 vessels
(c) 42 vessels
(d) 35 vessels
Ans: (b)
Total capacity of cistern = 60*15 = 900 liters
Total vessels needed to fill the cistern if the total volume of each vessel is reduces to 20 liters = 60*15 / 20 = 900/20 = 45 vessels

Example 2: There are two pipes, Pipe M and Pipe N. They can fill a container respectively in 9 hours and 3 hours. Both Pipe M and Pipe N are opened alternatively in every hour and pipe N is started after pipe M. Identify in how many hours the container will be overflowed?
(a) 7 hr
(b) 5 hr
(c) 6 hr
(d) 4 hr
Ans: (b)
Container filled by pipe M in 1 hour = 1 / 9
Container filled by pipe N in 1 hour = 1 / 3
Container filled by both if open alternatively = 1/9 + 1/3 = 4/9 (in 2 hours)
Container filled by both in another 2 hours = 4/9 + 4/9 = 8/9
Remaining 1/9th part will be filled by pipe M in another one hour.
Total time = 2+2+1 = 5 hours

Example 3: An electric pump is used for emptying and filling a storage chamber of water. The storage chamber contains 3200 liters of water. What is the filling capacity, if the emptying of the storage chamber is 20 liters/minute more than its filling capacity and the pump requires 8 minutes less to empty the storage chamber than it requires to fill it?
(a) 100
(b) 80
(c) 70
(d) 50
Ans: (b)
Let us assume the filling capacity of the pump be x liter/minute
The emptying capacity will be 20+x liter/minute
Time needed to fill the storage chamber = 3200 / x
Time needed to empty the storage chamber = 3200 / x+20
Given, the pump requires 8 minutes less in emptying the storage chamber,
= 3200/x – 3200/x+20 = 8
= 400/x – 400/x+20 = 1
= 400(x+20) – 400x = x(x+20)
= 400x + 8000 – 400x = x² + 20x
= 8000 = x² + 20x
= x² + 20x – 8000 = 0
= (x +  100) (x - 80) = 0
x = -100, x = 80
Since liters cannot be negative, therefore the filling capacity is 80 liters/ minute.

Example 4: A pipe normally fill a cistern in four hours less than 24 hours. Because of a crack on one side of the cistern the pipe will take 5 hours more to fill. If the cistern is completely filled, in how much time the leakage will completely unfill the cistern?
(a) 100 hrs
(b) 110 hrs
(c) 120 hrs
(d) 125 hrs
Ans: (a)
The cistern is filled in 20 hours
In 1 hour = 1/20^th portion of the cistern is filled
Due to crack the cistern is filled in 25 hours
In 1 hour = 1/25^th portion of the cistern is filled
Part of the tank emptied in a hour = 1/20 – 1/25 = 1/100
Therefore, the cistern will be completely unfilled in 100 hours.

Example 5: Due to an open outlet pipe a tank will tank 4 hours more to be filled. Normally it takes 8 hours to completely fill. How many hours the outlet pipe will take to empty the full tank?
(a) 20 hrs
(b) 24 hrs
(c) 18 hrs
(d) 19 hrs
Ans: (a)
The complete tank is filled in 8 hours, so in 1 hour = 1/8^th tank will be filled
Due to outlet pipe in 1 hour = 1/12^th part will be filled
Part of the tank emptied in 1 hour = 1/8 – 1/12 = 1/24
Therefore the tank will be emptied in 24 hours.