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Ratios and Dilutions

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Caogan looked for patterns in the mixtures first of all, using a trial and error to find the first few answers, and then proceed to generalise into a formula:

To make it simpler to record my results, first I decided to let $U$ be the volume of undiluted liquid added in, let $W$ be the volume of water added, and let $D$ be the concentraion of the new substance.

First, I did a few experiments
U (ml) W (ml) D (cells/ml) D (as a fraction of original)
100 100 50,000 $\frac{1}{2}$
50 100 33,333.$\dot{3}$ $\frac{1}{3}$
100 50 66,666.$\dot{6}$ $\frac{2}{3}$
10 30 25,000 $\frac{1}{4}$
20 20 50,000 $\frac{1}{2}$
30 10 75,000 $\frac{3}{4}$
10 40 20,000 $\frac{1}{5}$
20 30 40,000 $\frac{2}{5}$
30 20 60,000 $\frac{3}{5}$
40 10 80,000
$\frac{4}{5}$

So I found the half, third, quarter and fifth that the question asked for.
I have also shown that you can also make fractions with a numerator greater than 1. These are found when the amount of $U$ is not an expicit factor of the sum of $U$ and $W$
I found a formula using this discovery. I noticed that the fraction of the concentration of the new substance is exactly $\frac{U}{U+W}$ and then to find the concentration $D$ we multiply by the original concentration, and so
$D=\frac{U}{U+W} \times 100,000$

With this formula we can see why there are more than one way of making most concentrations. Anything where there are equivilances to the fraction, then there are more than one way of making it. For example
$\frac{1}{2} = \frac{10}{10+10} = \frac{50}{50+50} = \frac{20}{20+20}$
$\frac{1}{4} = \frac{10}{10+30} = \frac{20}{20+60} = \frac{50}{50+150}$
$\frac{2}{3} = \frac{20}{20+10} = \frac{30}{30+15} = \frac{40}{40+20}$

So the concentraions we can't make are any where they cannot be expressed as a fraction. I have slightly cheated above as I have allowed $15 ml$ of water to be used, and in the example we can only use $10, 20, ... , 90, 100 ml$. So the concentrations we can't make are anything such as $\frac{7}{7+5}$ which is $58,333.\dot{3}$

Caogan did well, using trial and error to find patterns, and then linking them using a formula. Can you think of any other places where this method can be useful?
Jen and Tess built on Caogan's formula, extending it to the two dilution problem:

In Caogan's formula, you have the proportions of $U$ and $W$, which is the fraction of the original strength that the new substance is, then multiplied by the concentration of the original.
So with two dilutions you must have the proportion of the strength of the first dilution, times the fraction of the strength of the original. So if we notate using:
$U$ - volume of undiluted solution
$W$ - volume of water added in first dilution
$D$ - volume of diluted solution
$V$ - volume of water added in second dilution
$C$ - concentraion of solution after two dilutions
Then we get
$C = \frac{U}{U+W} \times \frac{D}{D+V} \times 100,000$

With Jen and Tess's formula, can any one now calculate the given dilutions to make?
Archie continues the problem:

The number of ways to make a final concentration of 25,000 cells/ml will be the same as the number of ways of making the fraction $\frac{1}{4}$ from a product of two fractions.
For example
U (ml) W (ml) D (ml) V (ml) $\frac{U}{U+W} \times \frac{D}{D+V}$ C (fraction of 100,000)
50 50 50 50 $\frac{50}{50+50} \times \frac{50}{50+50}$ $\frac{1}{2}\times \frac{1}{2}$
40 60 50 30 $\frac{40}{40+60} \times \frac{50}{50+30}$ $\frac{2}{5} \times \frac{5}{8}$

I decided to see if I could come up with a formula for finding the concentration of a substance after arbritarily many dilutions. Using the pattern above, we have that the concentration after dilution will always be $\frac{volume.of.diluted.solution}{volume.of.diluted.solution+volume.of.water.added} \times concentration.of.solution$
So if we let $U_i$ be the volume of the solution used at step $i$, let $W_i$ be the volume of water used at step $i$, and let $C_i$ be the concentraion after i dilutions then the concentration after $n$ steps will be
$C_n=\prod_{i=1}^n (\frac{U_i}{U_i + W_i}) \times 100,000$

Excellent work Archie! Developing a formula for one, two and then many cases is extremely useful. Can anyone now use Archie's formula to answer the last few questions on what concentrations are possible?