Zoha’s class is raising money for a local charity. The class puts any money raised into a pouch, and each Thursday their teacher creates a math problem about the money in the pouch.
The following note was attached to the pouch today.
This pouch contains a total of \(\$20.30\) in Canadian money consisting of \(4\) coins and \(3\) bills.
What are the specific bills and coins in the pouch?
What is the solution to the problem? Justify your answer.
Note: The coins available in Canada are nickels that are worth \(5\) cents, dimes that are worth \(10\) cents, quarters that are worth \(25\) cents, loonies that are worth \(\$1\), and toonies that are worth \(\$2\). Also, \(\$1\) is equal to \(100\) cents. The lowest denominations of bills are worth \(\$5\), \(\$10\), and \(\$20\).
The pouch cannot include a \(\$20\) bill since there is only \(30\) cents more than \(\$20\), and that would mean the pouch only contained \(1\) bill. Similarly, it cannot include two \(\$10\) bills since this would mean the pouch only contained \(2\) bills.
If it has one \(\$10\) bill and two
\(\$5\) bills, then that would be a
total of \(\$20\). This is three
bills.
In this case, there are \(30\) cents
remaining, which can be formed by:
\(1\) quarter and \(1\) nickel for a total of \(2\) coins
\(3\) dimes for a total of \(3\) coins
\(2\) dimes and \(2\) nickels for a total of \(4\) coins
\(1\) dime and \(4\) nickels for a total of \(5\) coins
\(6\) nickels for a total of \(6\) coins
So one possibility is that the pouch contains one \(\$10\) bill, two \(\$5\) bills, two dimes, and two nickels. However, we should check to see if this is the only possibility.
Could it have three \(\$5\) bills which is \(\$15\)? This means there would be \(\$5.30\) remaining. The fewest number of coins you need to make \(\$5\) is two toonies and one loonie, which is a total of \(3\) coins. But you need at least \(2\) coins to make up \(30\) cents. So you need at least \(5\) coins to make \(\$5.30\), which is too many coins.
Any more attempts to come up with \(\$20\) would take more bills and coins. So the only possibility that meets the requirements of the problem is one \(\$10\) bill, two \(\$5\) bills, two dimes, and two nickels.