Difference between revisions of "2024 AMC 10B Problems/Problem 9"
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+ | ==Solution 3== | ||
+ | Assume that <math>a = 0</math> and <math>b = -c</math>. Plugging into the second equation, <math>b^{2} + b^{2} = 30</math>, so <math>b^2 = 15</math>. Observe that taking the positive or negative root won't matter, as c will be the opposite. If we let <math>b = \sqrt{15} and </math>c = -\sqrt{15}, <math>0\times\sqrt{15} + 0\times-\sqrt{15} + \sqrt{15}-\sqrt{15} is -15, and dividing by 3 gives us </math>\boxed{\textbf{(A)}\ -5}.$ | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 13:27, 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
Since we have and
From the second given, , so Substituting this into the above equation, Thus, and their arithmetic mean is
~laythe_enjoyer211, countmath1
Solution 3
Assume that and . Plugging into the second equation, , so . Observe that taking the positive or negative root won't matter, as c will be the opposite. If we let c = -\sqrt{15}, \boxed{\textbf{(A)}\ -5}.$
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/QLziG_2e7CY?feature=shared
~ Pi Academy
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.