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Difference between revisions of "1983 AHSME Problems/Problem 9"

(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...")
 
(Rewrote the solution using a simpler method, and increased clarity)
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==Solution==
 
==Solution==
  
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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, <math>\frac{34(11x)+32(10x)}{21x}</math>. Working this out, you get <math>\boxed{\textbf{(D)}\ 33\frac{1}{21}}</math>.
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Assume, without loss of generality, that there are exactly <math>11</math> women and <math>10</math> men. Then the total age of the women is <math>34 \cdot 11 = 374</math> and the total age of the men is <math>32 \cdot 10 = 320</math>. Therefore the overall average is <math>\frac{374+320}{11+10} = \boxed{\textbf{(D)}\ 33\frac{1}{21}}</math>.

Revision as of 17:39, 26 January 2019

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 $\frac{374+320}{11+10} = \boxed{\textbf{(D)}\ 33\frac{1}{21}}$.

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