<?xml version="1.0" encoding="UTF-8" ?>
  <resource>
  <id>278</id>
  <path>/www/nrich/html/content/99/09/15plus2/</path>
  <resourceTypeID>1</resourceTypeID>
  <last_published>2011-02-01T00:00:01</last_published>
  <indexXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;p&gt;Rudolff&amp;#39;s book, called Coss, written in 1525, was the first German algebra book. The reason for the title is that cosa is a &amp;#39;thing&amp;#39;, a term used for the unknown. Algebraists were called cossists, and algebra the cossic art, for many years. Here is the problem:&lt;br&gt;&lt;/br&gt;
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £ 2 for women and 50p for children. How many men, women and children are there in the group?&lt;/p&gt;&lt;/mdoxml&gt;</indexXML>
  <solutionXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;p&gt;Congratulations to Elizabeth Whitmore and Sue Liu of Madras
College, St Andrew's, Scotland, to Giulio Tiozzo, age 15, Liceo
Scientifico &amp;quot;Galileo Ferraris&amp;quot;, Turin, Italy, and to Vassil
Vassilev, age 14, Lawnswood High School, Leeds and Jonathan Kemp,
age 17, Westwood High School, Leek, England. Here is Jonathan's
solution.&lt;/p&gt;
&lt;p&gt;The problem states that a group of 20 people pay a total of
£20 to see an exhibition, admission £3 for
men, £2 for women and 50p for children.&lt;/p&gt;
&lt;p&gt;Denoting the number of men as $x$, the number of women as $y$
and the number of children$z$ you have to solve two
simultaneous equations:$$ x + y + z = 20$$ and $$3x +
2y + {z\over 2} = 20.$$ Multiplying the second equation by 2 and
subtracting the first equation to eliminate $z$ gives: $$ 5x + 3y
=20.$$ This is a Diophantine equation as featured in this months
article. It is easy to spot two solutions, namely $x = 4$ , $y =
0$, and $x = 1$, $y = 5$. In this case we have 2 simultaneous
equations and 3 unknowns.&lt;/p&gt;
&lt;p&gt;Substituting these solutions into the first two equations gives:
$ 4 + 0 + z = 20$ therefore $z = 16$ and $12 + 0 + z/2 =20$ and
again $z = 16$.&lt;/p&gt;
&lt;p&gt;Substituting $x = 1$ and $y = 5$ gives: $1 + 5 + z = 20$
therefore $z = 14$ and $3 + 10 + z/2 = 20$ and again $z = 14$.&lt;/p&gt;
&lt;p&gt;So the group can either consist of either 4 men, no women and 16
children, or of 1 man, 5 women and 14 children.&lt;/p&gt;
&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML/>
  <clueXML/>
  <canonXML/>
  <end_user_role>2</end_user_role>
  <difficulty>3</difficulty>
  <keystage1>0</keystage1>
  <keystage2>0</keystage2>
  <keystage3>0</keystage3>
  <keystage4>1</keystage4>
  <keystage4plus>0</keystage4plus>
  <title>Rudolff's Problem</title>
  <description>A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?</description>
  <spec_group>Algebra
    <specifier>Algebra - generally</specifier>
  </spec_group>
  <spec_group>Numbers and the Number System
    <specifier>Integers</specifier>
  </spec_group>
  <spec_group>Algebra
    <specifier>Diophantine equations</specifier>
  </spec_group>
  <spec_group>Algebra
    <specifier>Other equations</specifier>
  </spec_group>
  <spec_group>Algebra
    <specifier>Simultaneous equations</specifier>
  </spec_group>
</resource>