Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution 2)
 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
+
Small lights are hung on a string <math>6</math> inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of <math>2</math> red lights followed by <math>3</math> green lights. How many feet separate the 3rd red light and the 21st red light?
−
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.
+
'''Note:''' <math>1</math> foot is equal to <math>12</math> inches.
  
 
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5 </math>
 
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5 </math>
  
 
== Solution ==
 
== Solution ==
 +
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>.
  
−
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, so the answer is <math>\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}</math>
+
== 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.
 +
 
 +
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.
 +
 
 +
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>.
 +
 
 +
~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

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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_7&oldid=224850"