2015 AMC 10A Problems/Problem 2

Problem

A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$

Solution

Let $a$ be the amount of triangular tiles and $b$ be the amount of square tiles.

Triangles have $3$ edges and squares have $4$ edges, so we have a system of equations.

We have $a + b$ tiles total, so $a + b = 25$ .

We have $3a + 4b$ edges total, so $3a + 4b = 84$ .

Multiplying the first equation by $3$ on both sides gives $3a + 3b = 3(25) = 75$ .

Second equation minus the first equation gives $b = 9$ , so the answer is $\boxed{\textbf{(D) }9}$ .

Solution 2

If all of the tiles were triangles, there would be $75$ edges. This is not enough, so there need to be some squares. Trading a triangle for a square results in one additional edge each time, so we must trade out $9$ triangles for squares. Answer: $\boxed{\textbf{(D) }9}$

2015 AMC 10A (Problems • Answer Key • Resources)
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2015 AMC 10A Problems/Problem 2

Contents

Problem

Solution

Solution 2

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