4793
/www/nrich/html/content/id/4793/
1
2011-02-01T00:00:01
<mdoxml version="1.0"><br></br>
<mdo:image height="294" width="400" src="cans.jpg" alt="Two cans of unstirred pink paint"></mdo:image><br></br>
<br></br>
A decorator can buy pink paint from two manufacturers.<br></br>
<ul>
<li>Paint A is made up from red and white paint in the ratio
<span style="font-weight: bold;">$1:3$</span></li>
<li>Paint B is made up from red and white paint in the ratio
<span style="font-weight: bold;">$1:7$</span></li>
</ul>
<div>He can mix the paints to produce a different shade of
pink.</div>
<p>If Paint A and Paint B come in same size cans, what is the least
number he would need of each type in order to produce pink paint
containing red and white in the following ratios:</p>
<div style="margin-left: 6em;">
<ul>
<li>$1:4$</li>
<li>$1:5$</li>
<li>$1:6$</li>
</ul>
</div>
<div>Another decorator buys pink paint from two different
manufacturers:</div>
<ul>
<li>Paint C is made up from red and white paint in the ratio
<span style="font-weight: bold;">$1:4$</span></li>
<li>Paint D is made up from red and white paint in the ratio
<span style="font-weight: bold;">$1:9$</span></li>
</ul>
<p>What is the least number he would need of each type in order to
produce pink paint containing red and white in the following
ratios:</p>
<div style="margin-left: 6em;">
<ul>
<li>$1:5$</li>
<li>$1:6$</li>
<li>$1:7$</li>
<li>$1:8$</li>
</ul>
</div>
<div>Is it always possible to combine two paints made up in the
ratios <span style="font-weight: bold;">$1:x$</span> and
<span style="font-weight: bold;">$1:y$</span> and turn them into
paint made up in the ratio <span style="font-weight: bold;">$1:z$</span> ? (where $x < z < y$)</div>
<p>Experiment with a few more examples.</p>
<p>Can you describe an efficient way of doing this?</p>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=4794&part=">Mixing
More Paints</a> is a follow-up question to this one.<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
<p class="editorial">We received several <span style="text-decoration: underline;">incorrect</span> solutions like the
ones below:</p>
<p style="font-weight: bold;">Combining paints A (1:3) and B
(1:7):</p>
<br></br>
<table border="1">
<tbody>
<tr>
<td>
<div style="text-align: center;">Required</div>
<div style="text-align: center;">Ratio</div>
</td>
<td>
<div style="text-align: center;">Amount of<br></br>
paint A</div>
</td>
<td style="text-align: center;">Amount of<br></br>
paint B</td>
</tr>
<tr>
<td style="text-align: center;">1:4</td>
<td style="text-align: center;">3</td>
<td style="text-align: center;">1</td>
</tr>
<tr>
<td style="text-align: center;">1:5</td>
<td style="text-align: center;">1</td>
<td style="text-align: center;">1</td>
</tr>
<tr>
<td style="text-align: center;">1:6</td>
<td style="text-align: center;">1</td>
<td style="text-align: center;">3</td>
</tr>
</tbody>
</table>
<br></br>
<br></br>
<p><span style="font-weight: bold;">Combining paints C (1:4) and D
(1:9):</span></p>
<br></br>
<table border="1">
<tbody>
<tr>
<td>
<div style="text-align: center;">Required</div>
<div style="text-align: center;">Ratio</div>
</td>
<td>
<div style="text-align: center;">Amount of</div>
<div style="text-align: center;">paint C</div>
</td>
<td>
<div style="text-align: center;">Amount of</div>
<div style="text-align: center;">paint D</div>
</td>
</tr>
<tr>
<td style="text-align: center;">1:5</td>
<td style="text-align: center;">4</td>
<td style="text-align: center;">1</td>
</tr>
<tr>
<td style="text-align: center;">1:6</td>
<td style="text-align: center;">3</td>
<td style="text-align: center;">2</td>
</tr>
<tr>
<td style="text-align: center;">1:7</td>
<td style="text-align: center;">2</td>
<td style="text-align: center;">3</td>
</tr>
<tr>
<td style="text-align: center;">1:8</td>
<td style="text-align: center;">1</td>
<td style="text-align: center;">4</td>
</tr>
</tbody>
</table>
<br></br>
<p class="editorial">They are based on the misconception that you
can add the ratios to work out the necessary combinations. The
solutions given have assumed that the 'parts' in the ratios are of
equal size so that a can in the ratio 1:3 contains half the amount
of the one in the ratio 1:7.<br></br>
<br></br>
However, one can of paint A and one can of paint B does
<span style="text-decoration: underline;">not</span> produce paint
in the ratio 2:10 (or 1:5) as suggested above, since that would
assume that the one part red in can A has the same volume as the
one part red in can B.<br></br>
<br></br>
This can't be the case since there are 4 parts in can A and 8 parts
in can B,<br></br>
so 1/4 of can A is red and 1/8 of can B is red..<br></br>
<br></br>
To compare equal quantitities we will need to express the ratio of
the colours in can A as 2:6, so we have:<br></br>
in can A: 2/8 red and 6/8 white<br></br>
in can B: 1/8 red and 7/8 white<br></br>
<br></br>
Combining one can of each paint will now give us<br></br>
3/8 red and 13/8 white,<br></br>
that is, paint in the ratio 3:13<br></br>
<br></br>
Can anyone now solve this tough nut?<br></br>
<br></br>
Yes!<br></br>
<br></br>
We've received two correct solutions from Year 8 students at St
Albans High School.<br></br>
<br></br>
Click <a href="/content/id/4793/MixingPaintSolution1.pdf">here</a>
to see Anjali's solution and <a href="/content/id/4793/MixingPaintSolution2.pdf">here</a> to see Lydia's
solution.<br></br>
<br></br>
Well done Anjali and Lydia, and all the other students in Ms
Chapman's class who managed to crack this tricky problem.</p>
<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
A problem that challenges students to consider the effect of
combining ratios.<br></br>
<br></br>
The problem assumes that students understand ratios (they
understand that if $1/3$ is red it does not mean that the ratio of
red to white is $1:3$) and are able to simplify ratios.<br></br>
<br></br>
A possible start to using this problem in a classroom could involve
asking students to suggest possible combinations of paints A and B
(or C and D) and their suggestions could be written up for all to
see.<br></br>
<br></br>
It may be useful to provide students with red and white counters so
that they can simulate the effect of mixing the paints. However,
counters must be used with care; students need to understand that
they must have the same overall number of counters for each paint
mixture. <br></br>
<br></br>
Possible follow up questions:<br></br>
<ul>
<li>What is the ratio of red to white paint in these mixes?</li>
<li>Can any of these ratios be simplified?</li>
<li>Will any of these combinations produce the same shade of
pink?</li>
<li>Can you list the mixes in order as they become paler?</li>
</ul>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=4794&part=">Mixing
More Paints</a> is a follow-up question to this one.<br></br>
<br></br>
<br></br>
<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
Experiment with different combinations of paints A and B (and
paints C and D). <br></br>
<br></br>
The ratio <span style="font-weight: bold;">$1:4 = 2:8 = 3:12 =
4:16$</span> ...<br></br>
<br></br>
The ratio <span style="font-weight: bold;">$1:5 = 2:10 = 3:15 =
4:20$</span> ...<br></br>
<br></br>
B is whiter than A (and D is whiter than C), so what will be
special about the combinations that achieve the paler shades of
pink?<br></br>
<br></br>
What is the relationship between the ratios of the original paints
and the ratio of the mix?<br></br>
<br></br>
<br></br></mdoxml>
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Mixing Paints
A decorator can buy pink paint from two manufacturers. What is the
least number he would need of each type in order to produce
different shades of pink.
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