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166 lines
6.0 KiB
HTML
166 lines
6.0 KiB
HTML
<!DOCTYPE html>
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<html lang="en"><!-- #BeginTemplate "/Templates/Main.dwt" --><!-- DW6 -->
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<meta http-equiv="content-type" content="text/html; charset=UTF-8">
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<title>Puzzle - Pencils and Jars - Solution</title>
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<article id="content" role="main">
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<!-- #BeginEditable "Body" -->
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<h1 class="center">Pencils and Jars Solution</h1>
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<p class="center">A <b>solution </b>to the <a href="pencilsandjars.html">Pencils and Jars Puzzle</a></p>
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<p class="center"><img src="puzzles/images/pencils.html" alt="pencils" height="" width=""></p>
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<h2>The Puzzles</h2>
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<h3>Puzzle 1</h3>
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<div class="indent50px">
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<p>I have some pencils and some jars.<br>
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If I put 4 pencils into each jar I will have one jar left over.<br>
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If I put 3 pencils into each jar I will have one pencil left over.<br>
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How many pencils and how many jars?</p>
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</div>
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<p> </p>
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<h3>Puzzle 2</h3>
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<div class="indent50px">
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<p>Again I have some pencils and some jars<br>
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If I put 9 pencils into each jar I will have two jars left over<br>
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If I put 6 pencils into each jar I will have three pencils left over<br>
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How many pencils and how many jars?</p>
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</div>
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<p> </p>
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<h2>A Solution</h2>
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<p>Puzzle 1 can be solved simply by trying some
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different numbers.</p>
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<p>But if we want to solve Puzzle 2 it is a
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bit more tricky.</p>
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<p>So we will solve Puzzle 1 using <a href="algebra/index.html">Algebra</a>, to make it easier to tackle Puzzle 2.</p>
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<h3>Puzzle 1</h3>
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<p>The first stage is to read the question carefully:</p>
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<p class="center80">I have some pencils and some jars.<br>
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If I put 4 pencils into each jar I will have one jar left over<br>
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If I put 3 pencils into each jar I will have one pencil left over<br>
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How many pencils and how many jars?</p>
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<p>There are two distinct statements.</p>
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<div class="dotpoint">
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<p>When there are four pencils in each jar there will be one jar
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left over.</p>
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<p>We can
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say that <b>the number of jars is equal to the number
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of pencils divided by four plus another one.</b></p>
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<p>This can be expressed using algebra:</p>
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<p class="center"><b class="large">j = (p/4) + 1</b></p>
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</div>
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<p> </p>
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<div class="dotpoint">
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<p>When there are 3 pencils in each jar there will be one pencil
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left.</p>
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<p>We now need to calculate how many pencils there are.</p>
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<p>We can
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say that <b>the number of pencils will be equal to the number
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of jars times three plus another one.</b></p>
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<p>Using algebra again we have:</p>
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<p class="center"><b class="large">p = 3j + 1</b></p>
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</div>
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<p> </p>
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<p>We have ended up with two <a href="algebra/linear-equations.html">linear equations</a>.</p>
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<p>There are two ways
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to solve these: <a href="data/function-grapher.html">Graphically</a> or using <a href="algebra/systems-linear-equations.html">simultaneous equations</a>.
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We will use graphs!</p>
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<p>First rearrange the formula to make <b>j</b> the subject:</p>
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<p class="center"><b class="large">j = (p/4) + 1</b> (equation
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1)</p>
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<p class="center"><b class="large">j = (p − 1) ÷ 3</b> (equation 2)</p>
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr>
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<td>Plotting these on a graph using x=p and y=j;</td>
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<td><img src="images/graph1.gif" alt="graph 1" height="136" width="200"></td>
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</tr>
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<tr>
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<td>If we find the <b>intersection point</b> we have our answer</td>
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<td><img src="images/graph2.gif" alt="graph 2" height="136" width="200"></td>
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</tr>
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</tbody></table>
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<p class="center large">And that is it: there are 16 pencils and 5 Jars.</p>
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<h3> </h3>
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<h3>Puzzle 2</h3>
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<p>Ahh! That is for YOU to work out, using the same method I have just shown you.</p>
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<p> </p>
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<div class="related">
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<a href="puzzles/index.html">Puzzles Index</a>
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<a href="puzzle.html">Assorted Puzzles</a>
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</div>
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<!-- #EndEditable -->
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</article>
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<footer id="footer" class="centerfull noprint"></footer>
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<div id="copyrt">Copyright © 2022 Rod Pierce</div>
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