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

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

@@ Line 3: / Line 3: @@
 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>
+<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11</math>
 ==Solution==
@@ Line 16: / Line 15: @@
 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>.
+Multiplying the first equation by <math>3</math> on both sides gives <math>3a + 3b = 3(25) = 75</math>.
+Second equation minus the first equation gives <math>b = 9</math>, so the answer is <math>\boxed{\textbf{(D) }9}</math>.
+==Solution 2==
+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>
+==Solution 3==
+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>.
+==Video Solution (CREATIVE THINKING)==
+ https://youtu.be/ZuogVamZHjk
+~Education, the Study of Everything
-==Alternate Solution==
-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>
+==Video Solution==
+https://youtu.be/MNUTCkQ0c-g
+~savannahsolver
 ==See Also==
 {{AMC10 box|year=2015|ab=A|num-b=1|num-a=3}}
 {{MAA Notice}}

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