2025 AMC 8 Problems/Problem 9

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 (ProblemsAnswer KeyResources)
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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