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>.
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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 $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}}\ 9\qquad\textbf{(E)}\ 11$ (Error compiling LaTeX. Unknown error_msg)


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 $3a + 4b = 84$.

Solving gives, $a = 16$ and $b = 9$, so the answer is $\boxed{\textbf{(D) }9}$.