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?