Difference between revisions of "2025 AMC 8 Problems/Problem 9"

Latest revision as of 19:52, 31 July 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

We proceed with brute force. All of the pairs of opposite numbers on the clock are $(12,6)$ , $(1,7)$ , $(2,8)$ , $(3,9)$ , $(4,10)$ , $(5,11)$ , where order doesn't matter. The averages of each of these pairs are $9, 4, 5, 6, 7,$ and $8$ respectively, and the average these numbers is $\frac{9+4+5+6+7+8}{6}=\boxed{\textbf{(B)}~6.5}$

~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

You could also see that the average of, $2$ and $8$ , is $5$ , only one more that it's consecutive pair, $1$ and $7$ which is $4$ . Assuming that this is the pattern, we can find the average $4$ to $9$ (6 consecutive integers starting with $4$ ), and we that find our answer is $\boxed{\textbf{(B)}~6.5}$ .

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

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

@@ Line 1: / Line 1: @@
-The 2025 AMC 8 is not held yet. Please do not post false problems.
+== Problem ==
+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?
+<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>
+<math>\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9.5 \qquad \textbf{(E)}\ 12</math>
+== Solution 1 ==
+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
+== Solution 2 ==
+We proceed with [[brute force]]. All of the pairs of 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>, <math>(5,11)</math>, where order doesn't matter. The averages of each of these pairs are <math>9, 4, 5, 6, 7,</math> and <math>8</math> respectively, and the average these numbers is <math>\frac{9+4+5+6+7+8}{6}=\boxed{\textbf{(B)}~6.5}</math>
+~Bepin999
+== Solution 3 (most efficient) ==
+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
+You could also see that the average of, <math>2</math> and <math>8</math>, is <math>5</math>, only one more that it's consecutive pair, <math>1</math> and <math>7</math> which is <math>4</math>. Assuming that this is the pattern, we can find the average <math>4</math> to <math>9</math> (6 consecutive integers starting with <math>4</math>), and we that find our answer is <math>\boxed{\textbf{(B)}~6.5}</math>.
+== Video Solution 1 by SpreadTheMathLove ==
+https://www.youtube.com/watch?v=jTTcscvcQmI
+== Video Solution 2 ==
+[//youtu.be/VP7g-s8akMY?si=QWgBnVJLRi_J9Hox&t=667 ~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
+==Video Solution(Quick, fast, easy!)==
+https://youtu.be/fdG7EDW_7xk
+~MC
+== See Also ==
+{{AMC8 box|year=2025|num-b=8|num-a=10}}
+{{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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