1983 AHSME Problems/Problem 9

Revision as of 18:39, 26 January 2019 by Sevenoptimus (talk | contribs) (Rewrote the solution using a simpler method, and increased clarity)

Problem

In a certain population the ratio of the number of women to the number of men is $11$ to $10$. If the average (arithmetic mean) age of the women is $34$ and the average age of the men is $32$, then the average age of the population is

$\textbf{(A)}\ 32\frac{9}{10}\qquad \textbf{(B)}\ 32\frac{20}{21}\qquad \textbf{(C)}\ 33\qquad \textbf{(D)}\ 33\frac{1}{21}\qquad \textbf{(E)}\ 33\frac{1}{10}$

Solution

Assume, without loss of generality, that there are exactly $11$ women and $10$ men. Then the total age of the women is $34 \cdot 11 = 374$ and the total age of the men is $32 \cdot 10 = 320$. Therefore the overall average is $\frac{374+320}{11+10} = \boxed{\textbf{(D)}\ 33\frac{1}{21}}$.