6882
/www/nrich/html/content/id/6882/
1
2011-02-01T00:00:01
<mdoxml version="1.0">
<br></br>
<ul id="stemLinks">
<li><a href="http://nrich.maths.org/6163">Warm-up</a></li>
<li><a href="http://nrich.maths.org/6164">Try this next</a></li>
<li><a href="http://nrich.maths.org/6165">Think higher</a></li>
<li><a href="http://en.wikipedia.org/wiki/Concentration">Read: mathematics</a></li>
<li><a href="http://en.wikipedia.org/wiki/Chemical_reaction">Read: science</a></li>
<li><a href="http://nrich.maths.org/6637">Explore further</a></li>
</ul>
<div> </div>
<br></br>
<p><span style="font-style: italic;">Scientists often require solutions which are diluted to a particular concentration. In this problem, you can explore the mathematics of simple dilutions.</span><br></br>
<br></br>
Imagine you have a beaker containing a solution with a concentration of 100 000 cells per millilitre of liquid. You can transfer some of this solution into a second beaker, in multiples of 10ml, and add water in multiples of 10ml to dilute the solution.<br></br>
<br></br>
If you diluted 100ml of the original solution with 100ml of water, what would be the concentration, in cells/ml, of your new solution? Use the interactivity below to see if you are right.<br></br>
<br></br>
<a href="/content/id/6882/6164simp.swf">Full Screen Version</a><br></br>
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<mdo:flash height="300" width="550"><param name="movie" value="/content/id/6882/6164simp.swf" ></param><param name="flashplayerversion" value="8" ></param></mdo:flash><br></br>
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Investigate other dilutions that can be made, using the interactivity to check your ideas.<br></br>
<br></br>
Here are some <span style="font-weight: bold;">questions to consider:</span></p>
<br></br>
<ul>
<li>Can you make solutions which are half the strength of the original?<br></br>
One third of the strength? One quarter? One fifth? ...</li>
<li>What about fractions with a numerator greater than 1?</li>
<li>Are there any concentrations you can make in more than one way?</li>
<li>What can you say about the concentrations you <span style="font-weight: bold;">can't</span> make?</li>
</ul>
<br></br>
<div>A series of dilutions can be performed, where a solution is diluted, and then the resulting solution is also diluted.</div>
<br></br>
<p>Use the interactivity below to investigate the concentrations which result from two dilutions. Try to predict what will happen before clicking "Get dilution" to check you are right.<br></br>
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<a href="/content/id/6882/6164med.swf">Full Screen Version</a><br></br>
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<mdo:flash height="180" width="550"><param name="movie" value="/content/id/6882/6164med.swf" ></param><param name="flashplayerversion" value="8" ></param></mdo:flash><br></br>
<br></br>
Find two dilutions which give a final concentration of:</p>
<ul>
<li>$50000$ cells/ml</li>
<li>$33333.\dot3$ cells/ml</li>
<li>$75000$ cells/ml</li>
<li>$49000$ cells/ml</li>
<li>$24000$ cells/ml</li>
<li>$45000$ cells/ml</li>
<li>$26666.\dot6$ cells/ml</li>
</ul>
<p>How many different ways can you find to make a final concentration of $25000$ cells/ml?<br></br>
<br></br>
Find some concentrations which are impossible to create using two dilutions.<br></br>
How can you convince yourself that they are not possible?<br></br>
List the necessary criteria for deciding whether a concentration is possible or not.<br></br>
<br></br>
<span style="font-style: italic;">You may wish to try the problems</span> <a href="http://nrich.maths.org/6164&part=" style="font-style: italic;">Investigating the Dilution Series</a> <span style="font-style: italic;">and</span> <a href="http://nrich.maths.org/6165&part=" style="font-style: italic;">Exact Dilutions</a><span style="font-style: italic;">, which expand on the ideas in this
problem.</span><br></br>
</p>
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<mdoxml version="1.0"><br></br>
<span class="editorial">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:</span><br></br>
<br></br>
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.<br></br>
<br></br>
First, I did a few experiments
<table border="1">
<tbody>
<tr>
<td>U (ml)</td>
<td>W (ml)</td>
<td>D (cells/ml)</td>
<td>D (as a fraction of original)</td>
</tr>
<tr>
<td>100</td>
<td>100</td>
<td>50,000</td>
<td>$\frac{1}{2}$</td>
</tr>
<tr>
<td>50</td>
<td>100</td>
<td>33,333.$\dot{3}$</td>
<td>$\frac{1}{3}$</td>
</tr>
<tr>
<td>100</td>
<td>50</td>
<td>66,666.$\dot{6}$</td>
<td>$\frac{2}{3}$</td>
</tr>
<tr>
<td>10</td>
<td>30</td>
<td>25,000</td>
<td>$\frac{1}{4}$</td>
</tr>
<tr>
<td>20</td>
<td>20</td>
<td>50,000</td>
<td>$\frac{1}{2}$</td>
</tr>
<tr>
<td>30</td>
<td>10</td>
<td>75,000</td>
<td>$\frac{3}{4}$</td>
</tr>
<tr>
<td>10</td>
<td>40</td>
<td>20,000</td>
<td>$\frac{1}{5}$</td>
</tr>
<tr>
<td>20</td>
<td>30</td>
<td>40,000</td>
<td>$\frac{2}{5}$</td>
</tr>
<tr>
<td>30</td>
<td>20</td>
<td>60,000</td>
<td>$\frac{3}{5}$</td>
</tr>
<tr>
<td>40</td>
<td>10</td>
<td>80,000</td>
<td>
<div>$\frac{4}{5}$</div>
</td>
</tr>
</tbody>
</table>
<br></br>
<div>So I found the half,
third, quarter and fifth that the question asked for.</div>
<div>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 $W$</div>
<div>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</div>
<div>$D=\frac{U}{U+W} \times
100,000$</div>
<br></br>
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<br></br>
$\frac{1}{2} = \frac{10}{10+10} = \frac{50}{50+50} =
\frac{20}{20+20}$<br></br>
$\frac{1}{4} = \frac{10}{10+30} = \frac{20}{20+60} =
\frac{50}{50+150}$<br></br>
$\frac{2}{3} = \frac{20}{20+10} = \frac{30}{30+15} =
\frac{40}{40+20}$<br></br>
<br></br>
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 $58,333.\dot{3}$<br></br>
<br></br>
<span class="editorial">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?</span><br></br>
<span class="editorial">Jen
and Tess built on Caogan's formula, extending it to the two
dilution problem:</span><br></br>
<br></br>
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.<br></br>
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:<br></br>
$U$ - volume of undiluted solution<br></br>
$W$ - volume of water added in first dilution<br></br>
$D$ - volume of diluted solution<br></br>
$V$ - volume of water added in second dilution<br></br>
$C$ - concentraion of solution after two dilutions<br></br>
Then we get<br></br>
$C = \frac{U}{U+W} \times \frac{D}{D+V} \times 100,000$<br></br>
<br></br>
<span class="editorial">With
Jen and Tess's formula, can any one now calculate the given
dilutions to make?</span><br></br>
<span class="editorial">Archie
continues the problem:</span><br></br>
<br></br>
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.<br></br>
For example<br></br>
<table border="1">
<tbody>
<tr>
<td>U (ml)</td>
<td>W (ml)</td>
<td>D (ml)</td>
<td>V (ml)</td>
<td>$\frac{U}{U+W} \times \frac{D}{D+V}$</td>
<td>C (fraction of 100,000)</td>
</tr>
<tr>
<td>50</td>
<td>50</td>
<td>50</td>
<td>50</td>
<td>$\frac{50}{50+50} \times \frac{50}{50+50}$</td>
<td>$\frac{1}{2}\times \frac{1}{2}$</td>
</tr>
<tr>
<td>40</td>
<td>60</td>
<td>50</td>
<td>30</td>
<td>$\frac{40}{40+60} \times \frac{50}{50+30}$</td>
<td>$\frac{2}{5} \times \frac{5}{8}$</td>
</tr>
</tbody>
</table>
<br></br>
<a href="" class="control" onclick="tablealter(2)"></a>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
$\frac{volume.of.diluted.solution}{volume.of.diluted.solution+volume.of.water.added}
\times concentration.of.solution$<br></br>
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<br></br>
$C_n=\prod_{i=1}^n (\frac{U_i}{U_i + W_i}) \times 100,000$<br></br>
<br></br>
<span class="editorial">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?</span><br></br>
<br></br></mdoxml>
<mdoxml version="1.0"><h3>Why do this problem?</h3>
<div>This problem provides a context within which to explore fractions and proportionality. Seeking concentrations which can be made in different ways and justifying why some concentrations can't be made at all gives practice on working with equivalent fractions, ratios, factors and multiples.</div>
<h3>Possible approach</h3>
<div>Begin by introducing the idea of a solution with strength 100000 cells/ml. Make sure everyone is clear about what this means.</div>
<br></br>
<div>Then ask the class to imagine mixing 100ml of this solution with 100ml of pure water. Students could record any working out and their answer on individual whiteboards. Now show students the <a href="/content/id/6882/6164simp.swf">interactivity</a>, and use it to check their answer. Take time to discuss how they got to the correct answer.</div>
<br></br>
<div>Demonstrate that the interactivity can measure multiples of 10ml of liquid, up to 100ml - the scientific context of this could be using a dropper that measures 10ml at a time.</div>
<br></br>
<div>Ask students to come up with questions they would like to explore using the interactivity - some suggested questions appear in the problem. Then allow them some time to investigate, using the interactivity to check the predictions that they make.</div>
<br></br>
<div>Once students are competent at working with solutions created using one dilution, show them the second <a href="/content/id/6882/6164med.swf">interactivity</a>, which performs two dilutions. Perhaps start by giving them a couple of concentrations to work out, using individual whiteboards as before, and using the interactivity to check. At the end of the problem there are some suggested
concentrations they could be asked to make.</div>
<br></br>
<div>Pairs of students could take it in turns to create a concentration using two dilutions, and then challenge their partner to work out the dilutions they used.</div>
<br></br>
<div>Finally, the problem challenges students to investigate impossible dilutions.<br></br>
<br></br>
<em>You can read about <a href="https://www.ncetm.org.uk/resources/34329">one teacher's experience</a> of using this task in the classroom.</em><br></br>
</div>
<h3>Key questions</h3>
<div>If I combine $a$ ml of solution with $b$ ml of water, how does the concentration change?</div>
<div>What happens when several dilutions are performed one after another?</div>
<div>Does the order in which I do the dilutions matter?</div>
<h3>Possible extension</h3>
<div>The problem <a href="http://nrich.maths.org/6164&part=">Investigating the Dilution Series</a> uses the same context but explores four steps of dilution instead of just one or two. <a href="http://nrich.maths.org/6165&part=">Exact Dilutions</a> extends these same ideas further.</div>
<br></br>
<h3>Possible support</h3>
<p><a href="http://nrich.maths.org/6870&part=">Mixing Lemonade</a> investigates the strengths of different solutions informally and may provide a useful starting point.<br></br>
<br></br>
</p></mdoxml>
<mdoxml version="1.0">What proportion of the new solution is the old solution? What
proportion is water? <br></br>
Think about how you can use these proportions to express the
concentration of the new solution each time.<br></br>
<br></br></mdoxml>
<mdoxml version="1.0">A single fraction in its simplest form is possible iff numerator
< 11, denominator < numerator+11. <br></br></mdoxml>
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Ratios and Dilutions
Scientists often require solutions which are diluted to a
particular concentration. In this problem, you can explore the
mathematics of simple dilutions
Fractions, Decimals, Percentages, Ratio and Proportion
Calculating with fractions
Fractions, Decimals, Percentages, Ratio and Proportion
Ratio
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