6748
/www/nrich/html/content/id/6748/
1
2011-02-01T00:00:01
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
<p>Dan has nine £$2$ coins and his sister Ann has eight £$5$ coins. What is the smallest number of coins that must change hands so that Dan and Ann end up with equal amounts of money?<br></br>
<br></br>
If you liked this problem, <a href="http://nrich.maths.org/public/viewer.php?obj_id=2648&part=">here is an NRICH task</a> which challenges you to use similar mathematical ideas.</p>
<p> </p>
<p> </p></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
Suppose Dan gives Ann $x$ £$2$ coins and Ann gives Dan
$y$ £$5$ coins. Then Dan has £$(18-2x+5y)$
and Ann has £$(40-5y+2x)$.<br></br>
<br></br>
We need $18-2x+5y=40-5y+2x\Rightarrow 10y-4x=22\Rightarrow
5y-2x=11$. The solution to this equation which minimises $x+y$ is
$x=2$, $y=3$, so the smallest number of coins that must change
hands is $5$. <br></br></mdoxml>
2
3
0
1
1
0
0
Weekly Problem 48 - 2009
Measures and Mensuration
Money