Pipes and Cistern


  1. There are three taps of diameter 1 cm, 4/3 cm and 2 cm, respectively. The ratio of the water flowing through them is equal to the ratio of the square of their diameters. The biggest tap can fill the tank alone in 61 min. If all the taps are opened simultaneously, how long will the tank take to be filled?









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    Time taken to fill the tank by the tap having 2 cm diameter = 61 min

    ∴ Time taken to fill the tank by the tap having 1 cm diameter
    = 61 x (2/1)2 = 244 min

    Similarly, time taken to fill the tank by the tap having 4/3 cm diameter
    = 61 x [(2 x 3)/4]2 = 61 x 9/4 = 549/4 min.

    Correct Option: D

    Time taken to fill the tank by the tap having 2 cm diameter = 61 min

    ∴ Time taken to fill the tank by the tap having 1 cm diameter
    = 61 x (2/1)2 = 244 min

    Similarly, time taken to fill the tank by the tap having 4/3 cm diameter
    = 61 x [(2 x 3)/4]2 = 61 x 9/4 = 549/4 min.

    ∴ Part of the tank filled by all the three pipes in 1 min
    = 1/61 + 1/244 + 1/(549/4)
    = (36 + 9 + 16)/2196 = 61/2196 = 1/36
    Hence, required time taken = 36 min


  1. Two pipes can fill a cistern in 14 and 16 h, respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, it took 92 min more to fill the cistern. When the cistern is full, in what time will the leak empty it?









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    Part filled by 1st pipe in 1 h = 1/14
    Part filled by 2nd pipe in 1 h = 1/16
    Part filled by the two pipes in 1 h
    = ( 1/14 + 1/16 ) = (8 + 7)/112 = 15/112
    ∴ Time taken by these two pipes to fill the cistern = 112/15 h = 7 h 28 min
    Due to leakage, the time taken 7 h 28 min + 92 min = 9 h
    &thee4; Work done by ( two pipes + leak ) in 1 h = 1/9

    Correct Option: A

    Part filled by 1st pipe in 1 h = 1/14
    Part filled by 2nd pipe in 1 h = 1/16
    Part filled by the two pipes in 1 h
    = ( 1/14 + 1/16 ) = (8 + 7)/112 = 15/112
    ∴ Time taken by these two pipes to fill the cistern = 112/15 h = 7 h 28 min
    Due to leakage, the time taken 7 h 28 min + 92 min = 9 h
    &thee4; Work done by ( two pipes + leak ) in 1 h = 1/9
    Work done by the leak in 1 h = 1/9 - 15/112 = (112 - 135)/1008
    = - 23/1008
    ∴ Time taken by leak to empty the full cistern
    = 1008/23 = 4319/23 h



  1. Three taps A,B and C can fill a cistern in 10, 15 and 20 minutes respectively. They are all turned on at once, but after 3 minutes C is turned off. How many minutes longer will A and B take to fill the cistern ?









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    T = (10 x 15 x 20) / (10 x 15 + 10 x 20 + 15 x 20) = 60/13 minutes

    Correct Option: B

    T = (10 x 15 x 20) / (10 x 15 + 10 x 20 + 15 x 20) = 60/13 minutes
    Now, applying the given rule, we have [60/13 x y]/ [y - 60/13 + 3] = 20
    or y = 21/10 = 2 min. 6 seconds.


  1. A cistern can be filled by one of two pipes in 30 minutes and by the other in 36 minutes. Both pipes are opened together for a certain time but being particularly clogged only 5 / 6 of the full quantity of water flows through the former and only 9 / 10 through the later. The obstructions, however, being suddenly removed the cistern is filled in 151/2 minutes from that moment. How long was it before the full flow of water began ?









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    Net filling in last 151/2 minutes
    = 31/2 (1/30 + 1/36) = 341/360
    Now, suppose they remained clogged for x minutes.
    Net filling in these x minutes
    = (x/30 x 5/6 + x/36 x 9/10) = 19 x/360

    Correct Option: A

    Net filling in last 151/2 minutes
    = 31/2 (1/30 + 1/36) = 341/360
    Now, suppose they remained clogged for x minutes.
    Net filling in these x minutes
    = (x/30 x 5/6 + x/36 x 9/10) = 19 x/360
    Remaining part = (1 - 19x/360) = (360 - 19x/360)
    360 - 19x/360 = 341/360 or x = 1.
    Hence, the pipes remained clogged for 1 minutes.



  1. Pipe A fills the cistern in half an hour and pipe B in 40 minutes, but owing to a crack in the bottom of the cistern it is found that pipe A now takes, 40 minutes to fill the cistern. How long will B take now to fill it and how long will the crack take to empty it ?









  1. View Hint View Answer Discuss in Forum

    Let the leak empties it in T hours
    From the given rule, we have (T x 30) / (T - 30) = 40

    Correct Option: B

    Let the leak empties it in T hours
    From the given rule, we have (T x 30) / (T - 30) = 40
    ∴ T = 120 minutes = 2 hours.

    Now, from the question, applying the rule, we have time taken by B to fill the tank when crack in the bottom develops
    = (120 x 40) / (120 - 40) = 60 minutes = 1 hour