# Problem of the Week Problem D and Solution Small Change

## Problem

Carroll and Arthur cleaned their house and found a total of $$33$$ coins. The coins were either nickels ($$5$$ cent coins), dimes ($$10$$ cent coins), or quarters ($$25$$ cent coins). There were twice as many quarters as dimes, and the total value of all the coins they found was $$\5.25$$.

How many of each type of coin did they find?

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

## Solution

Let $$n$$ be the number of nickels, $$d$$ be the number of dimes, and $$q$$ be the number of quarters.

From the total number of coins we get the equation $n + d + q = 33 \tag{1}$

From the value of the coins we get the equation $5n + 10d + 25q=525 \tag{2}$

We also know that $$q = 2d$$.

Substituting $$q=2d$$ into equation $$(1)$$ and simplifying, we get \begin{aligned} n + d + 2d &= 33\end{aligned} $n + 3d = 33 \tag{3}$ Substituting $$q=2d$$ into equation $$(2)$$ and simplifying, we get $5n + 10d + 25(2d) = 525$ $5n + 60d = 525$ $n + 12d = 105 \tag{4}$ We can isolate $$n$$ in equation $$(3)$$ to get $$n = 33 - 3d$$.
Similarly, we can isolate $$n$$ in equation $$(4)$$ to get $$n = 105 - 12d$$.

Since $$n=n$$, it follows that \begin{aligned} 33 - 3d &=105 - 12d\\ -3d + 12d &= 105 - 33\\ 9d &=72\\ d &= 8 \end{aligned} Substituting $$d=8$$ into $$n = 33 - 3d$$, it follows that $$n=33-3(8) = 33-24 = 9$$.

Finally, we substitute $$d=8$$ into $$q=2d$$, to find $$q = 2(8) = 16$$.

Therefore, they found $$9$$ nickels, $$8$$ dimes, and $$16$$ quarters.