Problems on Pipes and Water Tank

Learn how to calculate the problems on pipes and water tank or cistern. We know, work done by inlet is positive and work done by outlet is negative.

Word problems on pipes and water tank or cistern:

1. A cistern can be filled by a tap in 12 hours and by the other tap in 9 hours. If both taps are opened together how long will it take to fill the cistern?

Solution:

Time taken by the 1st tap to fill the cistern = 12 hours

Therefore, work done by the 1 st tap in 1 hour = 1/12

Time taken by the 2nd tap to fill the cistern = 9 hours.

Therefore, work done by the 2nd tap in 1 hour = 1/9

Therefore, work done by the both taps in 1 hour = 1/12 + 1/9

                                                                 = (3 + 4)/36

                                                                 = 7/36

Therefore, both taps will fill the cistern in = 36/7 hours.

2. A pipe can fill the tank in 5 hours.  Due to leakage at the bottom it is filled in 6 hours. When the tank is full, in how much time will it be empties by the leak?

Solution:

When there is no leakage, the pipe can fill the cistern in 5 hours.

Therefore, The pipe fills 1/5 th part of the tank in one hour.

When there is a leakage, the pipe can fill the cistern in 6 hours.

In case of leakage, the pipe fills 1/6 th part of the tank in one hour.

So, in 1 hour due to leakage (1/5 – 1/6) th

                                      = (6 – 5)/30 th

                                      = 1/30 th

The part of the tank is emptied.

So, the tank will be emptied by leakage in 30 hours.

3. A tank can be filled by two tap A and B in 8 hour and 10 hours respectively. The full tank can be emptied by the third tap in 9 hours. If all the taps be turned on at the same time, in how much time will the empty tank be filled up completely?

Solution:

Time taken by tap A to fill the tank = 8 hours

Time taken by tap B to fill the tank = 10 hours

Time taken by tap C to fill the tank = 9 hours

Therefore, tap A fills 1/8 th part of the tank in 1 hour.

Tap B fills 1/10 th part of the tank in 1 hour.

Tap C empties out 1/9 th part of the tank in 1 hour.

Thus, in 1 hour (1/8 + 1/10 – 1/9) part of the tank is filled.

(45 + 36 – 40)/360 = 41/360 th part of the tank is filled.

Thus, tank will be filled completely in 360/41 hours, when all the three taps A, B and C are opened together.

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