2015 AMC 10A Problem 2

Start with an easy guess and fix it.

1. Pretend all $25$ tiles are triangles. Each triangle has $3$ edges, so that would be $25 \times 3 = 75$ edges.
2. The real count is $84$, which is $84 - 75 = 9$ edges more than the guess.
3. Now swap one triangle for a square. A square has $4$ edges, one more than a triangle, so each swap adds exactly $1$ edge and keeps the number of tiles at $25$.
4. To gain $9$ edges we need $9$ swaps. So there are $9$ squares (and $25 - 9 = 16$ triangles).
5. Check: $9$ squares give $9 \times 4 = 36$ edges, $16$ triangles give $16 \times 3 = 48$ edges, and $36 + 48 = 84$. Correct.

The answer is $\boxed{\textbf{(D)}\ 9}$.

This is the same puzzle as "chickens and cows with so many heads and legs". Assume everything is the cheaper kind, then each swap changes the total by a fixed amount, so the leftover divided by that amount counts the swaps. It is the same as solving $3(25 - s) + 4s = 84$, just without the letters.

Use a letter. Let $s$ be the number of squares, so there are $25 - s$ triangles. Counting edges,
$$3(25 - s) + 4s = 84, \quad\text{so}\quad 75 + s = 84, \quad\text{so}\quad s = 9.$$