Pipes and Cistern


  1. Two pipes A and B can fill a tank in 18 and 6 h, respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank ?









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    Part filled by A in 1 h = 1/18
    Part filled by B in by 1 h = 1/6
    Part filled (A + B) in 1 h =1/18 + 1/6 = (1 + 3)/18 = 4/18 = 2/9

    Correct Option: A

    Part filled by A in 1 h = 1/18
    Part filled by B in by 1 h = 1/6
    Part filled (A + B) in 1 h =1/18 + 1/6 = (1 + 3)/18 = 4/18 = 2/9
    Hence, both the pipes together will fill the tank in 9/2 h or 41/2 h


  1. A cistern can be filled up in 4 h by an inlet A. An outlet B can empty the cistern in 8 h. If both A and B are opened simultaneously, then after how much time will the cistern get filled?









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    Part filled by A in 1 h = 1/4
    Part emptied by B in 1 h = 1/8
    Part filled by (A + B) In 1 h = 1/4+ ( -1/8 ) = 1/4 - 1/8 = (2 - 1)/8 = 1/8

    Correct Option: C

    Part filled by A in 1 h = 1/4
    Part emptied by B in 1 h = 1/8
    Part filled by (A + B) In 1 h = 1/4+ ( -1/8 ) = 1/4 - 1/8 = (2 - 1)/8 = 1/8
    ∴ Required time to fill the cistern = 8



  1. Pipes A and B can fill a tank in 5 and 6 h, respectively. Pipe C can fill it in 30 h. If all the three pipes are opened together, then in how much time the tank will be filled up?









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    Part filled by A in 1 h = 1/5
    Part filled by B in 1 h = 1/6
    Part filled by C in 1 h = 1/30

    Correct Option: B

    Part filled by A in 1 h = 1/5
    Part filled by B in 1 h = 1/6
    Part filled by C in 1 h = 1/30
    Net part filled by ( A + B + C ) in h 1 = ( 1/5 + 1/6 + 1/30 )
    = (6 + 5 + 1)/30 = 12/30 = 2/5

    ∴ Required time to fill the tank = 5/2 = 21/2 h


  1. A tap can fill an empty tank in 12 h and a leakage can empty the tank in 20 h. If tap and leakage both work together, then how long will it take to fill the tank?









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    Part filled by tap in 1 h = 1/12
    Part emptied by leak in 1 h = 1/20
    Net part filled in 1 h when both (tap and leakage) work = 1/12 - 1/20
    = (5 - 3)/60 = 2/60 = 1/30

    Correct Option: C

    Part filled by tap in 1 h = 1/12
    Part emptied by leak in 1 h = 1/20
    Net part filled in 1 h when both (tap and leakage) work = 1/12 - 1/20
    = (5 - 3)/60 = 2/60 = 1/30

    ∴ Required time to fill the tank = 30 h



  1. Three taps A, B and C together can fill an empty cistern in 10 min . The tap A alone can fill it in 30 min and the tap B alone can fill it in 40 min. How long will the tap C alone take to fill it?









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    Part filled by (A + B + C ) in 1 min = 1/10
    Part filled by A in 1 min = 1/30
    Part filled by B in 1 min = 1/40
    Part filled by (A+B) in 1 min = 1/30 + 1/40 = (4 + 3)/120 = 7/120

    Correct Option: B

    Part filled by (A + B + C ) in 1 min = 1/10
    Part filled by A in 1 min = 1/30
    Part filled by B in 1 min = 1/40
    Part filled by (A+B) in 1 min = 1/30 + 1/40 = (4 + 3)/120 = 7/120
    ∴ Part filled by C in 1 min = 1/10 - 7/120 = (12 - 7)/120 = 5/120 = 1/24
    ∴ Tap C will fill the cistern in 24 min.