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

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==Solution==
 
==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, <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>.
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==See Also==
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{{AHSME box|year=1983|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 00:45, 20 February 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 $\frac{374+320}{11+10} = \boxed{\textbf{(D)}\ 33\frac{1}{21}}$.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions


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