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

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A box contains a collection of triangular and square tiles. There are <math>25</math> tiles in the box, containing <math>84</math> edges total. How many square tiles are there in the box?
 
A box contains a collection of triangular and square tiles. There are <math>25</math> tiles in the box, containing <math>84</math> edges total. How many square tiles are there in the box?
  
<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}}\ 9\qquad\textbf{(E)}\ 11</math>
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<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11</math>
 
 
  
 
==Solution==
 
==Solution==
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Let <math>a</math> be the amount of triangular tiles and <math>b</math> be the amount of square tiles.
 
Let <math>a</math> be the amount of triangular tiles and <math>b</math> be the amount of square tiles.
  
Triangles have 3 edges and squares have 4 edges, so we have a system of equations.
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Triangles have <math>3</math> edges and squares have <math>4</math> edges, so we have a system of equations.
  
 
We have <math>a + b</math> tiles total, so <math>a + b = 25</math>.
 
We have <math>a + b</math> tiles total, so <math>a + b = 25</math>.
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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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Multiplying the first equation by <math>3</math> on both sides gives <math>3a + 3b = 3(25) = 75</math>.
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Second equation minus the first equation gives <math>b = 9</math>, so the answer is <math>\boxed{\textbf{(D) }9}</math>.
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==Solution 2==
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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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==Video Solution==
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https://youtu.be/MNUTCkQ0c-g
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~savannahsolver
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==See Also==
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{{AMC10 box|year=2015|ab=A|num-b=1|num-a=3}}
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{{MAA Notice}}

Revision as of 18:01, 16 June 2020

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

Video Solution

https://youtu.be/MNUTCkQ0c-g

~savannahsolver

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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