# Ugh....need Big Help!

• Jul 27th 2005, 10:18 PM
Beth101
Ugh....need Big Help!
Hi,

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?

THANKS!
• Jul 28th 2005, 01:23 AM
Rebesques
what?
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 ( :confused: :rolleyes: :eek: :p ). 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. :mad:
• Jul 28th 2005, 01:56 AM
ticbol
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 --------******
where
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.

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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.

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).
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
Or,
1/x +1/y = 1/t -------the formula.

----------------------------------------
x = 9 min.
y = 15 min

So,
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.

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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.

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One big problem if you rely only on formulas, once you forget the formula then you are lost.