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Tricky Math Problem of the Week (11 Sep)

Solution to the 11 Sep, 2006 problem (Pipes And Cistern)

There are three taps A, B and C in a container. Tap A and B can fill the container in 15 and 10 minutes respectively, if they are working alone. Tap C can empty it in 12 minutes if it is working alone. Initially all the Taps are closed. Tap A is opened alone for 3 minutes. Now A is closed and B is opened alone for 2 minutes. Now B is closed and C is opened alone for 5 minutes. Finally all the three taps are opened together. How long will it take to fill the tank?

  • 10 minutes
  • 12 minutes
  • Tank Cannot be filled
  • None of these

    Answer :12 minutes

    Solution (Fast Approach)

    Part 1

  • Tap A alone can fill 100% of the container in 15 min.
    Its opened for 3 minutes. Hence it fills (3/15)x100 = 20% of the conatainer
  • In the same way in 2 minute B can fill (2/10)x100 = 20%.
  • Now container is 40% full
  • Now Tap C is opened for 5 min. It can empty (5/12)x100 = 41.67% of the container in 5 min. Only 40% of the container is filled, hence C will completely empty the container.
    Part 2
  • In 1 min A, B and C combined can fill (1/15 + 1/10 - 1/12) = 1/12 portion of the container
  • Hence 1 portion (Full) can be filled in 1/(1/12) = 12 minutes

  •  

    Tricky Math Problems of the Week

      23 Oct, 2006 Solution (Late Chauffer)

      16 Oct, 2006 Solution (Bee flying between two trains)

      9 Oct, 2006 Solution (Monkey Climbing a Rope)

      2 Oct, 2006 Solution (Bouncing Ball)

      25 Sep, 2006 Solution (Probability - Three Friends for the same job)

      18 Sep, 2006 Solution (Time and Work- Fencing Field)

      11 Sep, 2006 Solution (Pipes And Cistern)

      4 Sep, 2006 Solution (Triangle Property)

     
     
     

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