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Filling a tank. Two pipes are connected to the same tank. Working together they can fill the tank in 4 hrs. The larger pipe working alone can fill the tank 6 hrs less than the smaller one, HOw long would the smaller one take, working alone, to fill the tank?

  • math -

    Let time taken by smaller pipe be t hrs
    then time takes by larger pipe is t-6 hrs

    rate of smaller pipe = 1/t
    rate of larger pipe = 1/(t-6)
    combined rate = 1/t + 1/(t-6)
    = (2t-6)/(t(t-6))

    time at combined rate = 1/ [ (2t-6)/(t(t-6)) ]
    = t(t-6)/(2t-6)

    then t(t-6)/(2t-6) = 4
    t^2 - 6t = 8t - 48
    t^2 - 14t + 48 = 0
    (t-8)(t-6) = 0
    t = 8 or t = 6 , but t = 6 would make the rate of the larger pipe undefined, so we reject that answer


    So it would take 8 hours to fill with only the smaller pipe.

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