# Difference between revisions of "2015 AMC 10A Problems/Problem 2"

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We have <math>3a + 4b</math> edges total, so <math>3a + 4b = 84</math>. | We have <math>3a + 4b</math> edges total, so <math>3a + 4b = 84</math>. | ||

− | Solving gives, <math>a = 16</math> and <math>b = 9</math>, so the answer is <math>\boxed{\textbf{(D)}9}</math>. | + | Solving gives, <math>a = 16</math> and <math>b = 9</math>, so the answer is <math>\boxed{\textbf{(D) }9}</math>. |

## Revision as of 13:35, 4 February 2015

## Problem

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

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}}\ 9\qquad\textbf{(E)}\ 11$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

## Solution

Let be the amount of triangular tiles and be the amount of square tiles.

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

We have tiles total, so .

We have edges total, so .

Solving gives, and , so the answer is .