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Thread: Ugh....need Big Help!

  1. #1

    Ugh....need Big Help!


    I need help with a word problem: One pipe can fill a hot tub in 9 minutes, while a second pipe can fill it in 15 minutes. If the tub is empty, how long will it take both pipes together to fill the hot tub?

    I have several like this, so can you show me the formula so I can figure out how to plug the others in?

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  2. #2
    Super Member Rebesques's Avatar
    Jul 2005
    My house.


    Let's see... By the data of the problem, I suppose the pipes supply water at a constant rate. For the first pipe, call this rate $\displaystyle A$. If the pipe supplies $\displaystyle \alpha(t)$ units of water in time t minutes, then we have
    $\displaystyle \frac{d\alpha}{dt}(t)=A$, which integrates to $\displaystyle \alpha(t)=At+\alpha(0),$ the term $\displaystyle \alpha(0)$ being the initial supply at time $\displaystyle t=0$. I guess $\displaystyle \alpha(0)=0$, as we have to set the timer and open the tap simultaneously ( ). If the tub fills at $\displaystyle V$ units of water, then $\displaystyle \alpha(9)=9A=V$, and so $\displaystyle A=\displaystyle\frac{V}{9}$. We conclude that, for the first pipe, the supply is $\displaystyle \alpha(t)=\frac{V}{9}t$.

    Under the same suicidal considerations, we obtain that the supply of the second pipe is $\displaystyle \beta(t)= \frac{V}{15}t$. For both pipes, the supply is $\displaystyle \alpha(t)+\beta(t)=\frac{V}{9}t+\frac{V}{15}t=0.17 77Vt$ (almost). When the tub fills, we will have $\displaystyle 0.1777Vt=V$, from where $\displaystyle t=\frac{1}{0.1777}=5.625$ minutes...

    The tub is full. Now, lets give president Bush a bath.
    Last edited by Rebesques; Jul 28th 2005 at 01:26 AM.
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  3. #3
    MHF Contributor
    Apr 2005
    Here is one way.

    So you have several questions like that. That is because that is a popular question.
    And, I think there is a popular formula already developed for that type of question. I think it is
    1/x +1/y = 1/t --------******
    x = time the task can be done/finished alone by one performer.
    y = another time the task can be done/finished alone also by another performer.
    t = the time the task can be done/finished if the two performers perform together.

    We can derive that formula, why not.

    Let J = complete task to be done.
    and, performer A can finish it alone in x time.
    and, performer B can finish it also alone in y time.
    and, if A and B perform together, the J will be finished in t time.

    Like distance, task = rate*time
    So, rate = task/time ------***
    Rate of A alone is J/x.
    Rate of B alone is J/y.

    If A and B perform together, their combined rate is (J/x +J/y).
    task = rate*time
    J = (J/x +J/y)*t
    Divide both sides by t,
    J/t = J/x +J/y
    Divide both sides by J,
    1/t = 1/x +1/y
    1/x +1/y = 1/t -------the formula.

    Now, per your example,
    x = 9 min.
    y = 15 min

    1/x +1/y = 1/t
    1/9 +1/15 = 1/t
    Clear the fractions, multiply both sides by 9*15*t,
    1*15*t +1*9*t = 1*9*15
    15t +9t = 135
    24t = 135
    t = 135/24
    t = 5.625 min ---- the time the tub will fill up if the two pipes will do it together.

    Wait, there is a shorter formula, or a simplification of the derived formula above;

    1/x +1/y = 1/t
    ty +tx = x*y
    t = (x*y) / (x+y) -----shorter formula.

    t = (9*15)/(9+15) = 135/24 = 5.625 min----same.

    One big problem if you rely only on formulas, once you forget the formula then you are lost.
    Last edited by ticbol; Jul 28th 2005 at 02:04 AM.
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