Difference between revisions of "2012 AMC 12B Problems/Problem 7"
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== Problem == | == Problem == | ||
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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 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? | ||
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== Solution == | == Solution == | ||
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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 <math>\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}</math> | 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 <math>\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}</math> | ||
== See Also == | == See Also == | ||
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{{AMC12 box|year=2012|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2012|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:07, 19 October 2020
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?
Note: 1 foot is equal to 12 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. 
