Pipes & Cisterns: Solved Examples

Q1. Two pipes M and N can fill a tank in 20 hours and 30 hours respectively. If both pipes are opened together, then how much time it will take to fill the tank?
Sol: Part of the tank filled by M in one hour = 1/20, Part of the tank filled by N in one hour = 1/30. Hence part filled by (M + N) in I hour = 1/20 + 1/30 = 1/12. Hence both pipes can fill the tank together in 12 hours. OR direct formula can also be used as (20 × 30)/20 + 30 = 600/50 = 12 hours.
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Q2. A pipe can fill a tank in 40 hours. Due to a leak in the bottom, the tank is getting filled in 60 hours. In how much time the tank would be empty if leak is working alone?
Sol: Let us assume that leak empties the tank in x hours. So 1/40 – 1/x = 1/60. Solving this equation, we get x = 120. So the tank would be empty in 120 hours.
Q3. Two pipes A and B fill a tank in 30 minutes and 60 minutes respectively. A pipe C at the bottom can empty the tank in 120 minutes. If all three pipes were open simultaneously, how long does it take to fill the empty tank?
Sol: Pipe A fills 1/30th of the tank in a minute. Pipe B fills 1/60th of the tank in a minute. Pipe C drains 1/120th of the tank in a minute. Therefore, if all three are open then, (1/30 + 1/60 – 1/120)th of the tank will be filled in a min. i.e. 1/24 th of the tank will be filled in a minute. Therefore, the tank will be filled in 24 minutes.
Q4. Two filling pipes M and N can fill a tank in 20 hours and 60 hours respectively. There is an outlet P also. If all the pipes are opened together, then tank is full in 40 hours. How much time would be taken by P to empty the full tank if working alone?
Sol: Let us assume that leak empties the tank in x hours. So 1/20 + 1/60 – 1/x = 1/40. Solving this equation, we get x = 24. Hence the tank would be empty in 24 hours.
Q5: Two pipes A and B can fill a cistern in 32 and 48 minutes respectively. Both pipes being opened, find when pipe B must be turned off, so that the cistern gets filled in 24 minutes?
Sol: As the cistern is getting filled in 24 minutes, pipe A can fill only 24/32 = ¾ of the cistern in total time. This means the other ¼ must be filled by pipe B. Now B can fill the whole tank in 48 minutes, so ¼ of the tank can be filled in ¼ of 48 minutes i.e. 12 minutes. Now the pipe B is opened from the beginning, it should be turned off after 12 minutes and that is the answer. OR the standard approach can be applied. Let us assume that B is closed after x min. Then part filled by (A + B) in x min + part filled by A in (24 – x) min = 1. So x (1/32 + 1/48) + (24 – x) 1/32 = 1. Solving this equation, we get x = 12. So B should be closed after 12 min.
Q6. Pipes M and N can fill a tank in 60 minutes and 30 minutes respectively. Pipe P drains 24 litres of water in a minute. If all of them are kept open when the tank is full, the tank gets emptied in 60 minutes. How much water can the tank hold?
Sol. Let the drain pipe takes E minutes to empty the tank. Now 1/60 + 1/30 – 1/E = -1/60. The value on the Right Hand Side would be negative because full tank has been emptied. Solving this we get E = 15. If the drain pipe can empty a tank in 15 minutes and its drainage rate is 24 litres per minute, the capacity will be 15 x 24 = 360 litres.
Q7. Three pipes A, B and C can fill a cistern in 12 hours. After working at it for 6 hours, C is closed and A and B can fill it in 10 hours more. How many hours will C alone take to fill the cistern?
Sol. In 6 hours A, B and C would have filled 6 x 1/12 = 1/2 of the cistern. Remaining part is 1 – 1/2 = 1/2, which A and B have filled in 10 hours. → A and B can fill the whole cistern in 10 x 2/1 = 20 hours. Its given that A, B and C can fill the tank in 12 hours. So C can fill the cistern alone in 1/12 – 1/20 = 1/30 = 30. Hence C alone would take 30 hours to fill the tank.
Q8. If two pipes work simultaneously, the tank gets filled in 24 hours. One pipe takes 20 hours longer than the other. How many hours does faster pipe takes to fill the tank working alone?
Sol. Let us assume that faster pipe takes x hours to fill the tank working alone. Then the slower pipe takes (x + 20) hours to fill the tank working alone. When both of them are working together, it takes 24 hours to fill the tank. Hence 1/x + 1/(x +20) = 1/24. Solving this equation, we get x = 40. So the faster pipe takes 40 hours to fill the tank working alone.
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