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Math Help - Archimedes bath problem

  1. #1
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    Archimedes bath problem

    Could anyone help me with this problem please?

    Archimedes has a bath which is filled using three taps labelled A, B and C.

    The rates of flow of B is twice that of A, the rates of flow of C is three times that of A.

    Yesterday, Archimedes filled his bath as follows:

    Firstly he used all three taps for 2 hours. Then he turned off B and finished filling the bath in an additional 2 hours and 45 mins using taps A and C only (it's a very big bath).

    Today Archimedes is going to fill his bath (currently empty) using all three taps.

    How long will this take him?
    Last edited by Natasha1; November 22nd 2012 at 04:15 PM.
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  2. #2
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    Re: Archimedes bath problem

    let A = flow rate for tap A in units per hour

    (2 hrs)(A + 2A + 3A) + (2.75 hrs)(A + 3A) = 1 tub filled

    23A = 1

    A = (1 tub filled)/(23 hrs)


    using all 3 ...

    (A + 2A + 3A)t = 1 tub filled

    solve for t
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  3. #3
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    Re: Archimedes bath problem

    Hello, Natasha1!

    Archimedes has a bath which is filled using three taps labelled A, B and C.

    The rate of flow of B is twice that of A; the rate of flow of C is three times that of A.

    Yesterday, Archimedes filled his bath as follows:

    First, he used all three taps for 2 hours. Then he turned off B and finished filling the bath
    in an additional 2 hours and 45 mins using taps A and C only.

    Today Archimedes is going to fill his bath (currently empty) using all three taps.

    How long will this take him?

    Let a = number of hours for A to fill the bath alone.
    In one hour, A can fill \tfrac{1}{a} of the bath.

    Then B can fill the bath alone in \tfrac{a}{2} hours.
    In one hour, B can fill \frac{1}{\frac{a}{2}} = \tfrac{2}{a} of the bath.

    And C can fill the bath alone in \tfrac{a}{3} hours.
    In one hour, C can fill \frac{1}{\frac{a}{3}} = \tfrac{3}{a} of the bath.


    In 2 hours, the three taps can fill: 2(\tfrac{1}{a}) + 2(\tfrac{2}{a}) + 2(\tfrac{3}{a}) \,=\,\tfrac{12}{a} of the bath.

    In 2\tfrac{3}{4} = \tfrac{11}{4} hours, A and C can fill: \tfrac{11}{4}(\tfrac{1}{a}) + \tfrac{11}{4}(\tfrac{3}{a}) \,=\,\tfrac{11}{a} of the bath.

    These two quantities can fill the entire bath: . \tfrac{12}{a} + \tfrac{11}{a} \:=\:1

    . . Hence: . a \,=\,23


    A takes 23 hours to fill the bath.
    . . In one hour, A fills \tfrac{1}{23} of the bath.
    . . In x hours, A fills \tfrac{x}{23} of the bath.

    B takes \tfrac{23}{2} hours to fill the bath.
    . . In one hour, B fills \tfrac{2}{23} of the bath.
    . . In x hours, B fills \tfrac{2x}{23} of the bath.

    C takes \tfrac{23}{3} hours to fill the bath.
    . . In one hour, C fills \tfrac{3}{23} of the bath.
    . . In x hours, C fills \tfrac{3x}{23} of the bath.


    In x hours, the three taps will fill the entire bath.

    . . \frac{x}{23} + \frac{2x}{23} + \frac{3x}{23} \:=\:1\quad\Rightarrow\quad \frac{6x}{23} \:=\:1

    . . 6x \:=\:23 \quad\Rightarrow\quad x \:=\:\frac{23}{6}


    Therefore, the three taps will take 3\tfrac{5}{6} hours.
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