Difference between revisions of "2015 AMC 10A Problems/Problem 2"
Alexwin0806 (talk | contribs) m (→Solution) |
|||
| Line 23: | Line 23: | ||
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> | 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> | ||
| + | ==Video Solution== | ||
| + | https://youtu.be/MNUTCkQ0c-g | ||
| + | |||
| + | ~savannahsolver | ||
==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 17:01, 16 June 2020
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 
.
Multiplying the first equation by on both sides gives 
.
Second equation minus the first equation gives 
, so the answer is 
.
Solution 2
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:
Video Solution
~savannahsolver
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. 
