# 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.