#
Problem
of the Week

Problem
C and Solution

Gimme
Some Change

## Problem

Jean gave Karyna a bag of coins containing only nickels (\(5\) cent coins) and dimes (\(10\) cent coins). The total value of all
the coins in the bag was \(\$11\) and
there were \(16\) more nickels than
dimes in the bag.

How many coins in total were in the bag?

Note: In Canada, \(100\) cents is equal to \(\$1\).

## Solution

**Solution 1**

In this solution, we will solve the problem without using
algebra.

The bag had \(16\) more nickels than
dimes. These \(16\) nickels are worth
\(16\times 5=80\) cents, or \(\$0.80\). The remaining \(\$11.00-\$0.80=\$10.20\) would be made up
using an equal number of nickels and dimes. Each nickel-dime pair is
worth \(15\) cents, or \(\$0.15\). By dividing \(\$10.20\) by \(\$0.15\) we determine the number of
nickel-dime pairs that are required to make \(\$10.20\). Since \(\$10.20\div \$0.15=68\), we need \(68\) nickel-dime pairs. That is, we need
\(68\) nickels and \(68\) dimes to make \(\$10.20\). But there were \(16\) more nickels in the bag. Therefore,
there were a total of \(68+68+16=152\)
coins in the bag.

**Solution 2**

In this solution, we will solve the problem using algebra.

Let \(d\) represent the number of
dimes. Since there were \(16\) more
nickels than dimes in the bag, then there were \((d+16)\) nickels in the bag. Since each
dime is worth \(10\) cents, the value
of \(d\) dimes is \(10d\) cents.

Since each nickel is worth \(5\) cents, the value of \((d+16)\) nickels is \(5(d+16)\) cents. The bag contains a total
value of \(\$11\) or \(1100\) cents. Therefore, \[\begin{aligned}
\text{Value of Dimes (in cents)}+\text{Value of Nickels (in
cents)}&=\text{Total Value (in cents)}\\
10d+5(d+16)&=1100\\
10d+5d+80&=1100\\
15d&=1100-80\\
15d&=1020\\
d&=68\\
d+16&=84
\end{aligned}\] Therefore, there were \(68\) dimes and \(84\) nickels for a total of \(68+84=152\) coins in the bag.