Difference between revisions of "2024 AMC 12B Problems/Problem 11"
Kafuu chino (talk | contribs) m (→See also) |
|||
Line 11: | Line 11: | ||
Hence the mean is <math>\boxed{\textbf{(E) }\frac{91}{180}}</math> | Hence the mean is <math>\boxed{\textbf{(E) }\frac{91}{180}}</math> | ||
+ | ==Solution 2== | ||
+ | |||
+ | We can add a term <math>x_0</math> into the list, and the total sum of the terms won't be affected since <math>x_0=0</math>. Once <math>x_0</math> is added into the list, the average of the <math>91</math> terms is clearly <math>\frac{1}{2}</math>. Hence the total sum of the terms is <math>\frac{91}{2}</math>. To get the average of the original <math>90</math>, we merely divide by <math>90</math> to get <math>\frac{91}{180}</math>. Hence the mean is <math>\boxed{\textbf{(E) }\frac{91}{180}}</math> | ||
+ | |||
+ | ~tsun26 | ||
+ | |||
+ | |||
+ | Notice that $x_0+x_1+\cdots+x_{90} | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=B|num-b=10|num-a=12}} | {{AMC12 box|year=2024|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:59, 14 November 2024
Contents
[hide]Problem
Let . What is the mean of ?
Solution 1
Add up with , with , and with . Notice by the Pythagorean identity. Since we can pair up with and keep going until with , we get Hence the mean is
Solution 2
We can add a term into the list, and the total sum of the terms won't be affected since . Once is added into the list, the average of the terms is clearly . Hence the total sum of the terms is . To get the average of the original , we merely divide by to get . Hence the mean is
~tsun26
Notice that $x_0+x_1+\cdots+x_{90}
See also
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.