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2015 AMC 10A Problem 2

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2015 AMC 10A Problem 2AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

A box contains a collection of triangular and square tiles. There are 2525 tiles in the box, containing 8484 edges total. How many square tiles are there in the box?

  1. A)

    33

  2. B)

    55

  3. C)

    77

  4. D)

    99

  5. E)

    1111

Answer

D

Key insight

If all 25 tiles were triangles there would be 75 edges; each square adds exactly one extra edge, so 84 - 75 = 9 squares.

Solution

Let ss be the number of square tiles; then there are 25−s25 - s triangular tiles. Counting edges,
3(25−s)+4s=84⟹75+s=84⟹s=9.3(25 - s) + 4s = 84 \quad\Longrightarrow\quad 75 + s = 84 \quad\Longrightarrow\quad s = 9.

Check: 99 squares contribute 3636 edges and 1616 triangles contribute 4848, and 36+48=8436 + 48 = 84.

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

Why this works

Two unknowns with a fixed total is the classic "heads and legs" setup. Rather than solving a full system, imagine every tile is the cheaper kind (a triangle, 7575 edges) and note that swapping one triangle for a square raises the count by exactly one edge. The surplus 84−75=984 - 75 = 9 is then the number of swaps.

The trap

Solving for the number of triangles (16) and reporting it, or setting up 3t + 4s = 25 by mixing up counts and edges.

Common mistakes

  • Solving for the number of triangles (1616) and reporting it, or setting up 3t+4s=253t + 4s = 25 by mixing up counts and edges.
  • Arithmetic slips such as 3⋅25=653 \cdot 25 = 65, which shift the answer to a neighboring choice.

Techniques

Set up the equation/formula and compute; no special trick needed