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Difference between revisions of "2025 AMC 8 Problems/Problem 9"
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==Problem==
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== Problem ==
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Ningli looks at the 6 pairs of numbers directly across from each other on a clock. She takes the average of each pair of numbers. What is the average of the resulting 6 numbers?
 
  
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[[File:Amc8_2025_prob9.PNG]]
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Ningli looks at the <math>6</math> pairs of numbers directly across from each other on a clock. She takes the average of each pair of numbers. What is the average of the resulting <math>6</math> numbers?
  
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<math>\textbf{(A)}\ 5\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9.5 \qquad \textbf{(E)}\ 12</math>
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<asy>
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unitsize(1cm);
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draw(circle((0,0),2));
  
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==Solution 1==
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for(int i = 1; i <= 12; ++i)
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The answer can be expressed as (1+7)/2 + (2 + 8)/2 + ... + (6 + 12)/2, with the whole result divided by 6. Therefore, the answer of the question is the sum of the numbers from 1 through 12 divided by 2 * 6 = 12. The answer is 12/2 = 6 * 13 = 78/12, leading to the answer <math>\frac{78}{12}=\boxed{\textbf{(B)}~6.5}</math>.
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{
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draw(1.9*dir(90-i*30)-- 2*dir(90-i*30));//,linewidth(1pt)
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label("$"+string(i)+"$",2.3*dir(90-i*30));
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}
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draw(2*dir(-150)--2*dir(30),dashed);
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</asy>
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<math>\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9.5 \qquad \textbf{(E)}\ 12</math>
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== Solution 1 ==
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Our answer is <cmath>\frac{\frac{1+7}{2} + \frac{2+8}{2} + \cdots + \frac{6+12}{2}}{6} = \frac{\frac{1}{2}((1+7)+(2+8)+\cdots+(6+12))}{6} = \frac{1+2+3+4+5+6+7+8+9+10+11+12}{2 \cdot 6} = \frac{\frac{12 \cdot 13}{2}}{2 \cdot 6} = \frac{78}{12} =\boxed{\textbf{(B)}~6.5}</cmath>
  
 
~Sigmacuber
 
~Sigmacuber
  
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==Solution 2==
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== Solution 2 ==
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The pairs for the opposite numbers on the clock are <math>(12,6)</math>, <math>(1,7)</math>, <math>(2,8)</math>, <math>(3,9)</math>, <math>(4,10)</math>, and <math>(5,11)</math>. The averages of each of these pairs are <math>9, 4, 5, 6, 7,</math> and <math>8</math> respectively. The averages of <math>9, 4, 5, 6, 7,</math> and <math>8</math> are <math>\frac{39}{6}=\boxed{\textbf{(B)}~6.5}</math>
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The averages of all of the pairs for the opposite numbers on the clock are <math>(12,6)</math>, <math>(1,7)</math>, <math>(2,8)</math>, <math>(3,9)</math>, <math>(4,10)</math>, and <math>(5,11)</math>. The averages of each of these pairs are <math>9, 4, 5, 6, 7,</math> and <math>8</math> respectively. The averages of <math>9, 4, 5, 6, 7,</math> and <math>8</math> are <math>\frac{39}{6}=\boxed{\textbf{(B)}~6.5}</math>
  
 
~Bepin999
 
~Bepin999
  
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==Solution 3 (most efficient solution)==
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== Solution 3 (most efficient) ==
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If you read the problem carefully, you will find that it is basically asking for the average of all <math>12</math> numbers. To find the average of all <math>12</math> numbers, you add all the numbers from <math>1 - 12</math> which is <math>78</math>, and divide it by <math>12</math> because there are 12 terms, therefore, the answer is <math>\frac{78}{12}=\boxed{\textbf{(B)}~6.5}</math>.
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The problem is asking for the average of all <math>12</math> numbers. To find the average of all <math>12</math> numbers, you find the sum of all the integers from <math>1</math> to <math>12</math> which is <math>78</math>, and divide it by <math>12</math> because there are 12 terms. Therefore, the answer is <math>\frac{78}{12}=\boxed{\textbf{(B)}~6.5}</math>.
  
 
~JacQueen2024
 
~JacQueen2024
  
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==Video Solution 1 by SpreadTheMathLove==
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== Video Solution 1 by SpreadTheMathLove ==
 
https://www.youtube.com/watch?v=jTTcscvcQmI
 
https://www.youtube.com/watch?v=jTTcscvcQmI
  
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==Video Solution (A Clever Explanation You’ll Get Instantly)==
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== Video Solution 2 ==
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https://youtu.be/VP7g-s8akMY?si=QWgBnVJLRi_J9Hox&t=667
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[//youtu.be/VP7g-s8akMY?si=QWgBnVJLRi_J9Hox&t=667 ~hsnacademy]
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~hsnacademy
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== Video Solution 3 by Thinking Feet ==
  
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==Video Solution by Thinking Feet==
 
 
https://youtu.be/PKMpTS6b988
 
https://youtu.be/PKMpTS6b988
  
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==See Also==
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== Video Solution 4 by Cool Math Problems ==
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https://youtu.be/BRnILzqVwHk?si=Akl6WBBA3yIJYI4X&t=399
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== See Also ==
 +
 
 
{{AMC8 box|year=2025|num-b=8|num-a=10}}
 
{{AMC8 box|year=2025|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 19:17, 13 February 2025

Problem

Ningli looks at the $6$ pairs of numbers directly across from each other on a clock. She takes the average of each pair of numbers. What is the average of the resulting $6$ numbers?

[asy] unitsize(1cm); draw(circle((0,0),2)); for(int i = 1; i <= 12; ++i) { draw(1.9*dir(90-i*30)-- 2*dir(90-i*30));//,linewidth(1pt) label("$"+string(i)+"$",2.3*dir(90-i*30)); } draw(2*dir(-150)--2*dir(30),dashed); [/asy]

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9.5 \qquad \textbf{(E)}\ 12$

Solution 1

Our answer is \[\frac{\frac{1+7}{2} + \frac{2+8}{2} + \cdots + \frac{6+12}{2}}{6} = \frac{\frac{1}{2}((1+7)+(2+8)+\cdots+(6+12))}{6} = \frac{1+2+3+4+5+6+7+8+9+10+11+12}{2 \cdot 6} = \frac{\frac{12 \cdot 13}{2}}{2 \cdot 6} = \frac{78}{12} =\boxed{\textbf{(B)}~6.5}\]

~Sigmacuber

Solution 2

The averages of all of the pairs for the opposite numbers on the clock are $(12,6)$, $(1,7)$, $(2,8)$, $(3,9)$, $(4,10)$, and $(5,11)$. The averages of each of these pairs are $9, 4, 5, 6, 7,$ and $8$ respectively. The averages of $9, 4, 5, 6, 7,$ and $8$ are $\frac{39}{6}=\boxed{\textbf{(B)}~6.5}$

~Bepin999

Solution 3 (most efficient)

The problem is asking for the average of all $12$ numbers. To find the average of all $12$ numbers, you find the sum of all the integers from $1$ to $12$ which is $78$, and divide it by $12$ because there are 12 terms. Therefore, the answer is $\frac{78}{12}=\boxed{\textbf{(B)}~6.5}$.

~JacQueen2024

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution 2

~hsnacademy

Video Solution 3 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution 4 by Cool Math Problems

https://youtu.be/BRnILzqVwHk?si=Akl6WBBA3yIJYI4X&t=399

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 AJHSME/AMC 8 Problems and Solutions

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