Difference between revisions of "2012 AMC 12B Problems/Problem 7"

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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 it wants the answer in feet, so the answer is <math>\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}</math> since it wants the answer in feet.
 
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 it wants the answer in feet, so the answer is <math>\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}</math> since it wants the answer in feet.
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== See Also ==
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{{AMC12 box|year=2012|ab=B|num-b=6|num-a=8}}

Revision as of 22:51, 12 January 2013

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.

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5$

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 it wants the answer in feet, so the answer is $\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}$ since it wants the answer in feet.

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
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