# Problem on Pipes and Cisterns

• Difficulty Level : Medium
• Last Updated : 21 Aug, 2022

Question 1: 9 pumps working 8 hours a day can empty a reservoir in 20 days. How many such pumps needed to empty the same reservoir working 6 hours a day in 16 days.
Solution: Apply formula P1D1H1 /W1= P2D2H2/W2
9 x 20 x 8 = P2 x 16 x 6
P2= 15 pumps required

Question 2: A leak can empty a completely filled tank in 10 hours. If a tap is opened in completely filled tank which admits 4 liters of water per minute, then leak takes 15 hours to empty the tank. How many litres of water does the tank hold?
Solution: Take LCM (10, 15) = 30
Let leak pipe is A and A’s efficiency = 30/10 = 3

here combine efficiency = 2
So efficiency of the leakage (Pipe A) and another Pipe (Pipe B) which is filling the tank is= 30/15 = 2
Pipe A is emptying at 3 units/hour and Pipe B is filling using then emptying rate down to 2 units/hour.
So, filling pipe efficiency is 3 – 2 = 1 unit/ hour.
Pipe B will fill tank in 30/1=30 hours
Filling rate = 4 litre/minute
It will fill 4 x 60 = 240 litre/hour.
Total capacity= 240 x 30 = 7200 litres

Question 3: A, B and C can fill a tank in 6 hours together. After working for 2 hours, C is closed and A and B fill it in 7 hours more. Find the time taken by C alone to fill the tank?
Solution: Let the total capacity is 42 units.
(A + B + C) per hour work = 42/6 = 7 units.
They all worked for 2 hour.
Total water filled = 7 x 2 = 14 units.
Remaining capacity= 42 – 14 =28 units.
(A+B)’s efficiency= 28/7 = 4 units/hr
C’s efficiency = 7 – 4 =3 units/hr
C alone can fill the tank in 42/3 = 14 hours

Question 4: A tap drives at a rate of one drop/second and 800 drops = 100 ml. The number of litres water wasted in 30 days of a month is ?.
Solution: 1 sec = 1 drop
Number of seconds in 30 days= 30days x 24hrs x 60min x 60sec
Number of milli-liters wasted = (100 x 30days x 24hrs x 60min x 60sec)/800
= 324000 ml
Number of liter= 324000/1000
=324 liters water wasted.

Question 5: Two pipes A and B independently can fill a tank in 20 hours and 25 hours. Both are opened together for 5 hours after which the second pipe is turned off. What is the time taken by first pipe alone to fill the remaining portion of the tank?
Solution: Total unit water = LCM(20, 25) = 100 unit
A’s efficiency = 100/20 = 5 unit/hour
B’s efficiency =100/25 = 4 unit/hour
After 5 hour the water filled by A and B together = 5 x 9 =45 unit
Remaining unit = 100 – 45 = 55 unit
Time taken by A alone = 55/5 = 11 hours

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