Difference between revisions of "2008 AMC 8 Problems/Problem 10"
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The total of all their ages over the number of people is | The total of all their ages over the number of people is | ||
| − | <cmath>\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{\textbf{(D)}\ 34}</cmath> | + | <cmath>\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{\textbf{(D)}\ 34}.</cmath> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=9|num-a=11}} | {{AMC8 box|year=2008|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:22, 8 August 2021
Problem
The average age of the 
people in Room A is 
. The average age of the 
people in Room B is 
. If the two groups are combined, what is the average age of all the people?
Solution
The total of all their ages over the number of people is
![\[\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{\textbf{(D)}\ 34}.\]](http://latex.artofproblemsolving.com/8/5/d/85d4a6c33833905ed75fa3634d9bd8210083da51.png)
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
