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 $\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}$.

Solution 2

We know that both $3$ and $21$ are odd. This means that they both start their respective patterns of $rrggg$ as the first $r$ value.

Let's take a look at one full gap of two reds $rrgggr$, in this gap there is a total of $2.5$ feet of gap (considering that $6$ inches is half a foot). So for each gap of two reds there is a $2.5$ feet gap.

The total amount of red gaps of lights hung is $\frac{21-3}{2} = 9$ gaps. Multiplying $9 * 2.5 = 22.5$ inches. This gives $(E)$.

~PeterDoesPhysics

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AMC 12 Problems and Solutions

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