Difference between revisions of "2024 AMC 10B Problems/Problem 9"
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<math>= \frac{-15}{3} = \boxed{\textbf{(A) }-5}</math> | <math>= \frac{-15}{3} = \boxed{\textbf{(A) }-5}</math> | ||
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==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2024|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:35, 14 November 2024
Contents
[hide]Problem
Real numbers and have arithmetic mean 0. The arithmetic mean of and is 10. What is the arithmetic mean of and ?
Solution 1
If , that means , and . Expanding that gives . If , then . Thus, we have . Arithmetic will give you that . To find the arithmetic mean, divide that by 3, so
Solution 2
Given:
Square both sides to get:
Also given:
Substituting into equation ,
There are 3 terms, so the mean is:
~laythe_enjoyer211
See also
2024 AMC 10B (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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.