Difference between revisions of "2015 AMC 10A Problems/Problem 2"
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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. | + | 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>. | ||
Revision as of 22:55, 9 February 2015
Problem
A box contains a collection of triangular and square tiles. There are 
tiles in the box, containing 
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. Unknown error_msg)
Solution
Let 
be the amount of triangular tiles and 
be the amount of square tiles.
Triangles have 
edges and squares have 
edges, so we have a system of equations.
We have 
tiles total, so 
.
We have 
edges total, so 
.
Solving gives, 
and 
, so the answer is 
.
Alternate Solution
If all of the tiles were triangles, there would be 
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 
triangles for squares. Answer:
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
