Difference between revisions of "2015 AMC 10A Problems/Problem 2"
m (→Problem) |
|||
(11 intermediate revisions by 7 users not shown) | |||
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== | ||
Line 16: | Line 15: | ||
We have <math>3a + 4b</math> edges total, so <math>3a + 4b = 84</math>. | We have <math>3a + 4b</math> edges total, so <math>3a + 4b = 84</math>. | ||
− | + | Multiplying the first equation by <math>3</math> on both sides gives <math>3a + 3b = 3(25) = 75</math>. | |
+ | Second equation minus the first equation gives <math>b = 9</math>, so the answer is <math>\boxed{\textbf{(D) }9}</math>. | ||
− | == | + | ==Solution 2== |
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> | ||
+ | ==Solution 3== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==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}} |
Latest revision as of 17:11, 25 February 2021
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:
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
~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.