2013 AMC 10B Problem 6

You cannot simply average $11$ and $33$, because the two groups have different sizes. Go back to the totals instead.

1. Average $=$ total $\div$ number of people, so total $=$ average $\times$ number of people.
2. The $33$ fifth-graders have ages that add up to $33 \times 11 = 363$.
3. The $55$ parents have ages that add up to $55 \times 33 = 1815$.
4. All together there are $33 + 55 = 88$ people, with ages adding up to $363 + 1815 = 2178$.
5. The overall average is $\frac{2178}{88}$. Both numbers divide by $22$: $2178 \div 22 = 99$ and $88 \div 22 = 4$. So the average is $\frac{99}{4} = 24.75$.

Sense check: there are more parents than kids, so the average should be above the halfway point $22$ but below $33$. The value $24.75$ fits.

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

An average of two groups is pulled toward the bigger group. The safe method is always to rebuild the totals, add them, and divide by the total count. Averaging the two averages only works when the groups are the same size.

Only the ratio of group sizes matters: $33 : 55 = 3 : 5$. So imagine just $3$ kids aged $11$ and $5$ parents aged $33$. Their ages add up to $3 \times 11 + 5 \times 33 = 33 + 165 = 198$, and there are $8$ people, so the average is $\frac{198}{8} = 24.75$.

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 $\boxed{\textbf{(C)}\ 24.75}$.