2013 AMC 10B Problem 6

Recover the totals from the averages. The fifth-graders' ages sum to $33 \cdot 11 = 363$ and the parents' ages sum to $55 \cdot 33 = 1815$.

Combined, $33 + 55 = 88$ people have total age $363 + 1815 = 2178$. The overall average is
$$\frac{2178}{88} = \frac{2178 \div 22}{88 \div 22} = \frac{99}{4} = 24.75.$$

The answer is $\boxed{\textbf{(C)}\ 24.75}$.

Averages cannot be averaged unless the groups are the same size; the combined mean is the total of all ages divided by the total number of people. Equivalently, the answer is a weighted average of $11$ and $33$ with weights $\frac{33}{88} = \frac{3}{8}$ and $\frac{55}{88} = \frac{5}{8}$, which is why it lands closer to $33$.

Use the weights directly: $\frac{3}{8}\cdot 11 + \frac{5}{8}\cdot 33 = \frac{33 + 165}{8} = \frac{198}{8} = 24.75$. The ratio of group sizes $33:55 = 3:5$ is the only thing that matters.

• Averaging the two averages to get 22, ignoring that there are more parents than students.
• Dividing $2178$ by $33 + 33$ or another wrong count of people instead of $88$.