2010 AMC 10B Problems/Problem 4: Difference between revisions
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<cmath>\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}</cmath> | <cmath>\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}</cmath> | ||
==Video Solution== | |||
https://youtu.be/49jID-0tszU | |||
-Education, the Study of Everything | |||
==Video Solution== | |||
https://youtu.be/uAc9VHtRRPg?t=209 | |||
~IceMatrix | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=3|num-a=5}} | {{AMC10 box|year=2010|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | |||
Latest revision as of 15:59, 1 August 2022
Problem
For a real number 
, define 
to be the average of and 
. What is 
?
Solution
The average of two numbers, 
and 
, is defined as 
. Thus the average of and would be 
. With that said, we need to find the sum when we plug, 
, 
and 
into that equation. So:
![\[\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}\]](http://latex.artofproblemsolving.com/5/2/3/523076ad5f9af4ca536d8f70a8a1c7c6cc095b02.png)
Video Solution
-Education, the Study of Everything
Video Solution
https://youtu.be/uAc9VHtRRPg?t=209
~IceMatrix
See Also
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
