Difference between revisions of "2025 AMC 8 Problems/Problem 9"
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==Solution 1== | ==Solution 1== | ||
− | 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 | + | The answer can be expressed as <math>(1+7)/2 + (2 + 8)/2 + ... + (6 + 12)/2</math>, 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 <math>12/2</math> = <math>6 \times 13</math> = <math>78/12</math>, leading to the answer <math>\frac{78}{12}=\boxed{\textbf{(B)}~6.5}</math>. |
~Sigmacuber | ~Sigmacuber |
Revision as of 16:25, 2 February 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
The answer can be expressed as , 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
=
=
, leading to the answer
.
~Sigmacuber
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
Solution 3 (most efficient solution)
If you read the problem carefully, you will find that it is basically asking for the average of all numbers. To find the average of all
numbers, you add all the numbers from
which is
, and divide it by
because there are 12 terms, therefore, the answer is
.
~JacQueen2024
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=QWgBnVJLRi_J9Hox&t=667 ~hsnacademy
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.