2012 AMC 12B Problems/Problem 7

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 the question asks for the answer in feet, the answer is $\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}$ .

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

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2012 AMC 12B Problems/Problem 7

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