2008 AMC 8 Problems/Problem 10: Difference between revisions
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\textbf{(E)}\ 35</math> | \textbf{(E)}\ 35</math> | ||
==solution== | |||
(6*40+4*25)/(6+4)=340/10=34 | (6*40+4*25)/(6+4)=340/10=34 | ||
Revision as of 16:09, 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
(6*40+4*25)/(6+4)=340/10=34
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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