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

Revision as of 21:33, 10 January 2021

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

Solution 3

Video Solution

https://youtu.be/MNUTCkQ0c-g

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 22: / Line 22: @@
 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>
+==Solution 3==
 ==Video Solution==

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