Difference between revisions of "2025 AMC 8 Problems/Problem 9"
m (Protected "2025 AMC 8 Problems/Problem 9" ([Edit=Allow only administrators] (expires 17:59, 29 January 2025 (UTC)) [Move=Allow only administrators] (expires 17:59, 29 January 2025 (UTC)))) |
|||
| Line 1: | Line 1: | ||
| − | + | ==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? | ||
| + | |||
| + | <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== | ||
| + | Notice that this is simply the sum of the numbers on the clock face, all divided by <math>12</math>. Why? This is because for a given number, it will be divided by <math>2</math> due to its averaging with its counterpart on the other side of the clock face, and then it will be divided by <math>6</math> when we average out all the smaller averages. Since the sum of the first <math>12</math> positive integers is <math>78</math>, our answer is <math>\frac{78}{12}=\boxed{\textbf{(B)}~6.5}</math>. ~cxsmi | ||
Revision as of 20:56, 29 January 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?
Solution 1
Notice that this is simply the sum of the numbers on the clock face, all 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
