2012 AMC 12B Problem 7

Attempt Problem 7 of the 2012 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

Small lights are hung on a string 66 inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 22 red lights followed by 33 green lights. How many feet separate the 33rd red light and the 2121st red light?
Note: 11 foot is equal to 1212 inches.
1818
18.518.5
2020
20.520.5
22.522.5
Answer: E
Concepts:pattern recognitionunit conversion
Small hint:
Group the lights into repeating blocks of 55 (two red, three green)
Big hint:
The 33rd red light starts block 22 and the 2121st red light starts block 1111
Solution:
The lights repeat in blocks of 5,5, so consecutive blocks start 5⋅6=305\cdot6=30 inches, or 2.52.5 feet, apart.
Each block has one odd-numbered red light beginning it. The 33rd red light begins the 22nd block and the 2121st red light begins the 1111th block.
The distance between them is (11−2)⋅2.5=22.5(11-2)\cdot2.5=22.5 feet.
Thus, the correct answer is E.
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