1983 AHSME Problems/Problem 9

Revision as of 15:11, 27 June 2018 by Skyraptor79 (talk | contribs) (Created page with "== Problem== In a certain population the ratio of the number of women to the number of men is <math>11</math> to <math>10</math>. If the average (arithmetic mean) age of the...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

To find the average age of the population, you find the total sum of ages of both women and men and average the entire thing. So, $\frac{34(11x)+32(10x)}{21x}$. Working this out, you get $\boxed{\textbf{(D)}\ 33\frac{1}{21}}$.