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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?
  
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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>
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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>
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==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.
  
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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>.
  
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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>.
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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 needs 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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==Solution 3==
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Let <math>x</math> be the number of square tiles. A square has <math>4</math> edges, so the total number of edges from the square tiles is <math>4x</math>. There are <math>25</math> total tiles, which means that there are <math>25-x</math> triangle tiles. A triangle has <math>3</math> edges, so the total number of edges from the triangle tiles is <math>3(25-x)</math>. Together, the total number of edges is <math>4x+3(25-x)=84</math>. Solving our equation, we get that <math>x=9</math> which means that our answer is <math>\boxed{\textbf{(D) }9}</math>.
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/ZuogVamZHjk
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 +
~Education, the Study of Everything
 +
 
  
  
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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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==Video Solution==
 +
https://youtu.be/MNUTCkQ0c-g
  
 +
~savannahsolver
  
 
==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}}

Latest revision as of 22:07, 26 June 2023

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

Solution 3

Let $x$ be the number of square tiles. A square has $4$ edges, so the total number of edges from the square tiles is $4x$. There are $25$ total tiles, which means that there are $25-x$ triangle tiles. A triangle has $3$ edges, so the total number of edges from the triangle tiles is $3(25-x)$. Together, the total number of edges is $4x+3(25-x)=84$. Solving our equation, we get that $x=9$ which means that our answer is $\boxed{\textbf{(D) }9}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/ZuogVamZHjk

~Education, the Study of Everything



Video Solution

https://youtu.be/MNUTCkQ0c-g

~savannahsolver

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
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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