Difference between revisions of "2015 AMC 10A Problems/Problem 2"
(→Problem) |
|||
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? | 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) | + | <math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11</math> |
− | |||
==Solution== | ==Solution== |
Revision as of 19:45, 1 March 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?
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.