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 23:55, 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. ! Extra }, or forgotten $.)
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 |
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