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 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.