Difference between revisions of "2025 AMC 8 Problems/Problem 9"
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==Problem== | ==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? | 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]] | ||
<math>\textbf{(A)}\ 5\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9.5 \qquad \textbf{(E)}\ 12</math> | <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 2== | ==Solution 2== | ||
| − | The pairs for the opposite | + | 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 | ||
| + | |||
| + | ==Video Solution 1 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| + | |||
| + | ==Video Solution by Thinking Feet== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:43, 30 January 2025
Contents
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?
Solution 1
Notice that this is simply the sum of the numbers on the clock face divided by 
. Why? This is because for a given number, it will be divided by 
due to its averaging with its counterpart on the other side of the clock face, and then it will be divided by 
when we average out all the smaller averages. Since the sum of the first positive integers is 
, our answer is 
. ~cxsmi
Solution 2
The pairs for the opposite numbers on the clock are 
, 
, 
, 
, 
, and 
. The averages of each of these pairs are 
and 
respectively. The averages of and are 
~Bepin999
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Thinking Feet
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
