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>
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<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 $6$ people in Room A is $40$. The average age of the $4$ people in Room B is $25$. If the two groups are combined, what is the average age of all the people?

$\textbf{(A)}\ 32.5 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 33.5 \qquad \textbf{(D)}\ 34\qquad \textbf{(E)}\ 35$

Solution

The total of all their ages over the number of people is

\[\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{\textbf{(D)}\ 34}.\]

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
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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