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This problem is from Gelfand's Algebra (I'm self-studying, this is not homework):

A swimming pool is divided into 2 equal sections. Each sections has its own water supply pipe. To fill one section (using its pipe), you need $a$ hours. To fill the other sections you need $b$ hours. How many hours would you neeed if you turn on both pipes and remove the wall dividing the pool into sections?

-What I tried and understand (please read this because I can solve the problem already one way so I'm not just looking for a solution. I can't solve this problem "the other way" and what I need is a conceptual clarification as to why it isn't working that other way):

If I can find the time it would take both pipes together to fill up one section (1/2 pool), then I could multiply this rate by 2 and get the time it'd take to fill up the whole pool.

The first pipe fills one section in $a$ hours, which I write as a ratio $\frac{1}{a}$. The second pipe's rate is then $\frac{1}{b}$, so that working together they will fill up a single section at a rate of $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$.

Therefore, to fill up the entire pool will take (edited from Henning's correction) $2 \cdot \frac{ab}{a+b}$ hours.

-My problem:

I tried solving the problem using ratios of $\frac{a}{1}$ and $\frac{b}{1}$ instead, namely "$a$ hours per pool" and "$b$ hours per pool"; I'm sure there's a way to make it work but I can't seem to make sense out of it.

The rates are already over the same denominator, namely 1 pool section. Obviously, just adding them and ending up with $\frac{a+b}{1}$ is wrong, but I'm not sure why, or how to make this problem work thinking about it with the rates this way.

I would really appreciate a conceptual explanation (some words) on top of the algebra.

Thank you.

EDIT(motivated by Henning's comment): When I say I'm sure there's a way to solve the problem with the inverse rates, I mean that I don't see why it wouldn't be possible; so if it is indeed impossible, my question would become "what makes it impossible?"

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    $\begingroup$ I'm not sure there is a "why" other than you seem to have no valid reason to expect that to work. When you propose a calculation the initial onus is on you to explain why one would ever think that could work and you don't seem to have done so for $\frac a1+\frac b1$ ... $\endgroup$ – Henning Makholm May 8 '15 at 20:54
  • $\begingroup$ @HenningMakholm, good point. I guess I think there should be no difference between the two in terms of how they can be used to solve problems. In terms of graphing the rates, one rate and its inverse would be the same with a switch between axes, which seems very arbitrary. So I guess if you know it's impossible to solve the problem using the inverse rates, my question would become "why is it impossible?" or "what makes it impossible?" $\endgroup$ – jeremy radcliff May 8 '15 at 20:56
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    $\begingroup$ Incidentally, $2\frac{a+b}{ab}$ is not the correct answer -- if $a=b$ then certainly the time should be the same no matter whether there's a wall or not -- that is, $a$ hours -- but in that case $2\frac{a+b}{ab}$ reduces to $\frac 4a$. You should have divided the total size by the rate $\frac{a+b}{ab}$ instead of multiplying them. $\endgroup$ – Henning Makholm May 8 '15 at 21:01
  • $\begingroup$ @HenningMakholm, yes thank you. I edited it and mentioned your comment. $\endgroup$ – jeremy radcliff May 8 '15 at 21:05
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    $\begingroup$ Still not good. For a concrete example of the $a=b$ case, suppose that either hose can fill one pool in $100$ hours. Your answer $\frac12\cdot\frac{a+b}{ab}$ would still work out to $\frac 12\cdot\frac{200}{10,000}=\frac1{100}$ hour -- but clearly the time doesn't drop from 100 hours to less than a minute just because you remove the dividing wall. $\endgroup$ – Henning Makholm May 8 '15 at 21:09
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Judging from your first comment in reply to your comment:

There should be no difference between the two in terms of how they can be used to solve problems.

I think what you want to know is: "If I have two quantities $a$ and $b$ expressed in some (units), and the result I am looking for is proportional to $a+b$, what do I need to do, working with $1/a$ and $1/b$ expressed in (units$)^{-1}$ to get to the same result?".

The key difference between the two approaches is that generally $1/a +1/b \not= 1/(a+b)$. What you are looking for is a way to pair up the two inverses so that they give you the correct quantity. Basically a function of two variables $f(x,y)$ such that it satisfies the following:

\begin{equation} f\left(\frac{1}{x},\frac{1}{y}\right) = \frac{1}{x+y} \end{equation}

for all $x,y > 0$. You can convince yourself that the function \begin{equation} f(x,y) = xy \left( x+y \right)^{-1} \end{equation} does just that. Working in this "inverse space" is a bit more complicated, but still possible.

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  • $\begingroup$ Yes thank you, this is very helpful, it's what I was trying to get at. $\endgroup$ – jeremy radcliff May 9 '15 at 4:29
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The best computation I can come up with where one thinks in terms of "how long does it take to fill one pool" would be to imagine that the wall is still there, and water from the pool with the faster hose just spills over into the other once the faster pool is full.

Assume (without loss of generality) that $a$ is the faster hose, and Let $T$ be the time it takes to fill one pool with two hoses. In that case, filling both pools will take time $2T$. On the other hand, filling both pools will also take time $a$ plus the time it takes for both hoses to fill the part of the other pool that is still left by the time the water starts spilling over (which is $\frac{b-a}{b}$ of one pool). So $$ 2T = a+T\frac{b-a}{b}$$ and we can apply algebra to that to solve for $T$ and get $T=\frac{ab}{a+b}$. The time it takes to fill both pools is then $2T=\frac{2ab}{a+b}$.

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So one pipe has a "timescale" of $a$ hours per section, the other $b$ hours per section. After a given time, adding the contributions from both pipes, gives you how many sections have been filled:

\begin{equation*} \mbox{sections} = \frac{\mbox{time}}{\mbox{timescale}_1} + \frac{\mbox{time}}{\mbox{timescale}_2} \end{equation*}

Following the algebra will give you the result.

Alternatively, if you wanted to work with the rates $1/a$ and $1/b$ in sections per hour, you would add the contributions this way:

\begin{equation} \mbox{sections} = \mbox{time} \times \mbox{rate}_1 + \mbox{time} \times \mbox{rate}_2 \end{equation}

and you get the same result since rate = timescale$^{-1}$.

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