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
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 |
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