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September 18th, 2014, 02:11 AM   #1
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A question

Hello everyone,

I have a mathematical problem and I am not very good at math…

So here is the problem:

I’m a biology student, working on fern anatomy (does that not sound absolutely awesome?) and I have to infiltrate a tissue sample with resin. This hardens the material and allows you to cut it and view it under the microscope. The concentration of resin in your solution has to increase from 0% to 100% gradually, because otherwise the resin does not go into the tissue deep enough. You do this in the following steps: 0%, 10%, 20%, 30% 50%, 70%, 90% and 100%. (So a few steps are skipped and it is not a continuous increase of 10%).

The way in which you do this is by removing a certain volume of your solution and replacing it with the same volume of 100% resin. The volume of my solution is 1500 ul. The first step is simple: replace 10% of the solution with resin, so you have a 90% solution. But then: when you do the next replacement you remove a bit of the resin that is in your solution, namely: 0,1*the amount of solution you are replacing. Right? So you cannot simply add the same volume of without taking the loss of some of the resin from step 1 into account.

I understand that it is impossible to reach 100% resin with this method mathematically, without replacing 100% of your solution, but there has to be a practical solution to my problem.

It is not a massively important problem (and potentially way to easy for you all) and I don’t really need your answers to continue my research. But I know there has to be a way to calculate this and it bothers me that I don’t know how to.

Thanks in advance. I am looking forward to your answers.
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September 18th, 2014, 04:19 AM   #2
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Suppose we have a solution that is $\frac{a}{100}$ths ($a%$) resin and we want to make the solution $\frac{b}{100}$ resin. Further, suppose that we have $V$mL of solution.

Our inital solution contains $\frac{aV}{100}$mL of resin and the rest is water. We want to replace $S$mL of solution with resin, so that the final amount of resin is $\frac{bV}{100}$mL.

Thus we remove $\frac{aS}{100}$ of resin and add $S$mL.

We therefore want \begin{align*}\frac{aV}{100} - \frac{aS}{100} + S &= \frac{bV}{100} \\ S(100-a) &= V(b-a) \\[12pt] \implies S &= V \frac{b-a}{100-a} \end{align*}
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September 18th, 2014, 05:27 AM   #3
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Well... I tried something and I think I did get quite close to that (but quite close is never really right, right?). This makes more sense. Thank you for your answer

Have a nice day!
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