Difference between revisions of "2015 AMC 10A Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
− | If all of the tiles were triangles, there would be <math>75</math> edges. This is not enough, so there | + | If all of the tiles were triangles, there would be <math>75</math> edges. This is not enough, so there needs 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> |
==Solution 3== | ==Solution 3== | ||
− | Let x be the number of square tiles. A square has <math>4</math> edges, so the total number of edges from the square tiles is <math>4x</math>. There are <math>25</math> total tiles, which means that there are <math>25-x</math> triangle tiles. A triangle has <math>3</math> edges, so the total number of edges from the triangle tiles is <math>3(25-x)</math>. Together, the total number of | + | Let <math>x</math> be the number of square tiles. A square has <math>4</math> edges, so the total number of edges from the square tiles is <math>4x</math>. There are <math>25</math> total tiles, which means that there are <math>25-x</math> triangle tiles. A triangle has <math>3</math> edges, so the total number of edges from the triangle tiles is <math>3(25-x)</math>. Together, the total number of edges is <math>4x+3(25-x)=84</math>. Solving our equation, we get that <math>x=9</math> which means that our answer is <math>\boxed{\textbf{(D) }9}</math>. |
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+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/ZuogVamZHjk | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
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==Video Solution== | ==Video Solution== |
Latest revision as of 23:07, 26 June 2023
Contents
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 needs 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:
Solution 3
Let be the number of square tiles. A square has
edges, so the total number of edges from the square tiles is
. There are
total tiles, which means that there are
triangle tiles. A triangle has
edges, so the total number of edges from the triangle tiles is
. Together, the total number of edges is
. Solving our equation, we get that
which means that our answer is
.
Video Solution (CREATIVE THINKING)
https://youtu.be/ZuogVamZHjk
~Education, the Study of Everything
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.