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

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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>.
 
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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==Alternate Solution==
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If all of the tiles were triangles, there would be <math>75</math> 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 <math>9</math> triangles for squares. Answer: <math>\boxed{\textbf{(D) }9}</math>
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=1|num-a=3}}
 
{{AMC10 box|year=2015|ab=A|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:51, 5 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. ! Extra }, or forgotten $.)


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 $\boxed{\textbf{(D) }9}$.


Alternate Solution

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: $\boxed{\textbf{(D) }9}$


See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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