Difference between revisions of "2013 AMC 12B Problems/Problem 5"

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{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #5]] and [[2013 AMC 10B Problems|2013 AMC 10B #6]]}}
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==Problem==
 
==Problem==
 
The average age of <math>33</math> fifth-graders is <math>11</math>. The average age of <math>55</math> of their parents is <math>33</math>. What is the average age of all of these parents and fifth-graders?
 
The average age of <math>33</math> fifth-graders is <math>11</math>. The average age of <math>55</math> of their parents is <math>33</math>. What is the average age of all of these parents and fifth-graders?

Revision as of 11:39, 7 April 2013

The following problem is from both the 2013 AMC 12B #5 and 2013 AMC 10B #6, so both problems redirect to this page.

Problem

The average age of $33$ fifth-graders is $11$. The average age of $55$ of their parents is $33$. What is the average age of all of these parents and fifth-graders?

$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28$

Solution

The sum of the ages of the fifth graders is $33 * 11$, while the sum of the ages of the parents is $55 * 33$. Therefore, the total sum of all their ages must be $2178$, and given $33 + 55 = 88$ people in total, their average age is $\frac{2178}{88} = \frac{99}{4} = \boxed{\textbf{(C)}\ 24.75}$.

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 12 Problems and Solutions