Difference between revisions of "2012 AMC 12B Problems/Problem 7"
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== Solution == | == 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 <math>\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}</math> | + | We know the repeating section is made of <math>2</math> red lights and <math>3</math> 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 <math>44</math> lights in between the 3rd and 21st red light, translating to <math>45</math> <math>6</math>-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>. |
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| + | == Solution 2 == | ||
| + | We know that both <math>3</math> and <math>21</math> are odd. This means that they both start their respective patterns of <math>rrggg</math> as the first <math>r</math> value. | ||
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| + | Let's take a look at one full gap of two reds <math>rrgggr</math>, in this gap there is a total of <math>2.5</math> feet of gap (considering that <math>6</math> inches is half a foot). So for each gap of two reds there is a <math>2.5</math> feet gap. | ||
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| + | The total amount of red gaps of lights hung is <math>\frac{21-3}{2} = 9</math> gaps. Multiplying <math>9 * 2.5 = 22.5</math> inches. This gives <math>(E)</math>. | ||
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| + | ~PeterDoesPhysics | ||
== See Also == | == See Also == | ||
{{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}} | ||
Latest revision as of 00:46, 10 August 2024
Contents
Problem
Small lights are hung on a string 
inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 
red lights followed by 
green lights. How many feet separate the 3rd red light and the 21st red light?
Note: 
foot is equal to 
inches.
Solution
We know the repeating section is made of red lights and 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 
lights in between the 3rd and 21st red light, translating to 
-inch gaps. Since the question asks for the answer in feet, the answer is 
.
Solution 2
We know that both and 
are odd. This means that they both start their respective patterns of 
as the first 
value.
Let's take a look at one full gap of two reds 
, in this gap there is a total of 
feet of gap (considering that inches is half a foot). So for each gap of two reds there is a feet gap.
The total amount of red gaps of lights hung is 
gaps. Multiplying 
inches. This gives 
.
~PeterDoesPhysics
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. 
