Difference between revisions of "2015 AMC 10A Problems/Problem 2"
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Solving gives, <math>a = 16</math> and <math>b = 9</math>, so the answer is <math>\boxed{\textbf{(D) }9}</math>. | Solving gives, <math>a = 16</math> and <math>b = 9</math>, so the answer is <math>\boxed{\textbf{(D) }9}</math>. | ||
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+ | ==Alternate Solution== | ||
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+ | 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> | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2015|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:51, 5 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 3 edges and squares have 4 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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