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Pipes and Cisterns: Solved Questions | Quantitative Aptitude for SSC CGL PDF Download

Inlet

A conduit utilized for introducing water into a tank, classified as positive work done.

Outlet

A conduit employed for draining a water-filled tank, regarded as negative work done.

Pipes and Cisterns: Solved Questions | Quantitative Aptitude for SSC CGL

Rules for Pipes and Cisterns

  • Inlet is referred to as the pipe which is supposed to fill the tank.
  • A pipe that is meant to empty the tank is called as outlet.
  • When a pipe works for x hours for filling up the cistern, then 1/x portion of the cistern will be completed in 1 hour.
  • When a pipe works for y hours to empty the cistern, then the portion of the cistern emptied in 1 hour is equals to = 1/y
  • Pipe A might fill a cistern n times as quickly as another pipe B. This states that if slower pipe B works for x min to fill the empty tank,
    then quicker pipe A works for x/n min to fill up the empty cistern. If they work collaboratively, then the portion of the cistern which is loaded up in 1 hour isPipes and Cisterns: Solved Questions | Quantitative Aptitude for SSC CGL
  • In some scenarios the efficiency of the pipes would have been given such as in percentage, to find the actual rate multiply the efficiency percentage with rate.
  • Be very careful when using Units make sure all the units are same, or else answer might get wrong.

Formulas to keep in mind.

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 = x+ 20x
= 8000 = x+ 20x
= x+ 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 = 1/20th portion of the cistern is filled
Due to crack the cistern is filled in 25 hours
In 1 hour = 1/25th 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 = 1/8th tank will be filled
Due to outlet pipe in 1 hour = 1/12th 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.

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