2018 AMC 10A Problem 8

The $10$-cent coins are described by comparing them with the $5$-cent coins, so use the $5$-cent count as the unknown.

1. Name the counts. Let $x$ be the number of $5$-cent coins.
- $10$-cent coins: $x + 3$.
- $25$-cent coins: whatever is left of the $23$ coins, $23 - x - (x + 3) = 20 - 2x$.
2. Write the total value. In cents,
$$5x + 10(x + 3) + 25(20 - 2x) = 320.$$
3. Solve. Multiply out: $5x + 10x + 30 + 500 - 50x = 320$. Collect terms: $530 - 35x = 320$, so $35x = 210$ and $x = 6$.
4. Find every count. $6$ five-cent coins, $9$ ten-cent coins, and $20 - 12 = 8$ twenty-five-cent coins.
5. Check. Coins: $6 + 9 + 8 = 23$. Value: $30 + 90 + 200 = 320$ cents. Both match.
6. Answer the question asked. It wants how many more $25$-cent coins than $5$-cent coins: $8 - 6 = 2$.

The answer is $\boxed{\textbf{(C)}\ 2}$.

There are three unknown counts, but two of the facts ("$3$ more" and "$23$ in all") let you write two counts in terms of the third right away. Then the value fact becomes one equation with one unknown. Pick as your unknown the thing everything else is compared to, and reread the question at the end, since it asks for a difference, not a count.

Guess and check on the $5$-cent coins. Try $x = 5$: $8$ ten-cent coins and $10$ quarters, worth $25 + 80 + 250 = 355$ cents, too much. Each extra $5$-cent coin also adds a $10$-cent coin and removes two quarters, a change of $5 + 10 - 50 = -35$ cents. From $355$ we need to drop $35$, so $x = 6$, giving $8$ quarters and a difference of $2$.