Difference between revisions of "2012 AMC 12B Problems/Problem 7"
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== Problem == | == Problem == | ||
| − | Small lights are hung on a string 6 inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 2 red lights followed by 3 green lights. How many feet separate the 3rd red light and the 21st red light? | + | Small lights are hung on a string <math>6</math> inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of <math>2</math> red lights followed by <math>3</math> green lights. How many feet separate the 3rd red light and the 21st red light? |
| − | '''Note:''' 1 foot is equal to 12 inches. | + | '''Note:''' <math>1</math> foot is equal to <math>12</math> inches. |
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5 </math> | <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5 </math> | ||
Revision as of 06:36, 29 June 2023
Problem
Small lights are hung on a string 
inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 
red lights followed by 
green lights. How many feet separate the 3rd red light and the 21st red light?
Note: 
foot is equal to 
inches.
Solution
We know the repeating section is made of 2 red lights and 3 green lights. The 3rd red light would appear in the 2nd section of this pattern, and the 21st red light would appear in the 11th section. There would then be a total of 44 lights in between the 3rd and 21st red light, translating to 45 6-inch gaps. Since the question asks for the answer in feet, the answer is 
See Also
| 2012 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
