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in this problem there are two different mathematical operations
which can be used: You can dilute a dilution, which corresponds to
multiplying fractions; you can mix two dilutions, which corresponds
to averaging two fractions.
Part 1:
1) To create a solution of $\frac{1}{3}$, pipette 10ml of solution
then 2 x 10ml of water into the 100ml beaker.
2) To create a solution of $\frac{1}{7}$, pipette 10ml of solution
followed by 6 x10ml of water into the 100ml beaker.
3) To create a solution of $\frac{1}{18}$ first create a solution
of $\frac{1}{9}$ by pipetting 10ml of solution followed by 8 x 10ml
of water into the 100ml beaker. Then, pippette 10ml of this
solution into another beaker, followed by 10ml of water to give the
required concentration.
4) To create a solution of $\frac{5}{12}$ first create a solution
of $\frac{5}{6}$ by pipetting 5 x 10ml of solution followed by 10ml
of water into the 100ml beaker. Then pipette 10ml of this solution
into the second 100ml beaker, followed again by 10ml of
water.
5) To create a solution of $\frac{29}{35}$ first note that
$\frac{29}{35}=\frac{2}{5}+\frac{3}{7}$. This means that
$\frac{29}{35}$ is the average of $\frac{4}{5}$ and $\frac{6}{7}$,
so our strategy is to mix equal amounts of these solutions. First
make a $\frac{4}{5}$ solution in beaker A by pipetting 4 x 10ml of
solution followed by 10ml of water. Then make a $\frac{6}{7}$
solution in beaker B by pipetting 6 x 10ml of solution followed by
10ml of water. Finally, pipette 10ml of the solution in beaker A
and 10ml of the solution in beaker B into beaker C, creating 20ml
of a $\frac{29}{35}$ solution.
6) To create a solution of $\frac{71}{882}$ requires a few steps.
First, create a $\frac{4}{7}$ solution by pipetting 4 x 10ml into
the beaker, followed by 3 x 10ml water. Then pippette out 10ml of
this into another beaker, followed by 6 x 10ml of water, to give a
solution of $\frac{4}{49}$. Discard the original $\frac{4}{7}$
solution and create a $\frac{5}{7}$ solution by pipetting 5 x 10ml
solution then 2 x 10ml water. Pipette 10ml of this out into the
third beaker, then add 8 x 10 ml of water to give a $\frac{5}{63}$
solution. Now, mix equal amounts of this solution with the
$\frac{4}{49}$ solution to give the required
$\frac{71}{882}$.
NB there is more than one way to make each solution!
Part 2:
The concentrations of $\frac{1}{2}$ to $\frac{1}{10}$ are trivially
created by adding 10ml of solution, followed by (n-1) x 10ml of
solution, where n is the denominator of the concentration to be
created.
For dilutions less than this, it is not possible to create those
which have a denominator which is a PRIME number $p$. Indeed, it is
not possible to multiply together two numbers (i.e dilute a
dilution) to give this number, and averaging two fractions (i.e
mixing two existing dilutions) with denominators which aren't
divisible by $p$ gives a fraction whose denominator is not
divisible by $p$.
To create the dilutions which are non-prime involves diluting an
already made solution. For example, creating $\frac{1}{18}$
involves taking the $\frac{1}{9}$ solution and mixing it in a 1:1
ratio with water.
To see whether it is possible to create a dilution of
$\frac{1}{362880}$ it is best to express 362880 as a produce of
primes, and to see whether any of these primes are greater than 10.
If so, then the solution cannot be made, but if not, then it can be
made readily.
$362880 = 2^9 \times 5 \times 7 \times 9^2$. Thus it is possible to
make!
Part 3:
Common sense tells us that by mixing a solution of concentration
$c_A$ and $c_B$ we will obtain a solution with an intermediate
concentration. It can be seen that we will not obtain a more
concentrated solution than either, since there are insufficient
moles being added for this to be possible! Also, to get a more
dilute solution, water would need to be added.
This can be proved as follows:
Suppose I mix $a$ units of $A$ with $b$ units of $B$.
Using the general concentration formula I get
$$\frac{ac_A+bc_B}{a+b} = c_C$$
I can rearrange this to give $$ c_A = \frac{(a+b)c_C-bc_B}{a}
$$
Next note that $c_A> c_B$, so I have $$
\frac{(a+b)c_C-bc_B}{a}> c_B $$
This reduces to $c_C> c_B$ (using positivity of and b, of
course)
Below is a spreadsheet for the creation of the $\frac{1}{13}$
dilution. Notice how tiny the error is after only 10 repeats of
steps 3 and 4!!!
You might be wondering how this actually works. Overall we have
placed 10ml of solution among 120ml of water, which is the required
concentration. We have then essentially mixed it all together as
much as possible until the concentration is constant between both
beakers, which will give an overall concentration of
$\frac{1}{13}$.
To get a concentration of $\frac{1}{23}$ requires the use of the
third beaker. Again, we must add 10ml of solution to 220ml of
water. This can be facilitated by having beaker A and B full of
water, and Beaker C having 10ml of solution and 20ml of water. B is
used to fill C, and then A is used to fill B. After this C is used
to fill A, and the cycle repeated. Below is a spreadsheet for 10
iterations: