1983 AHSME Problems/Problem 9

Revision as of 23:45, 19 February 2019 by Sevenoptimus (talk | contribs) (Added box at the bottom)
(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

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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png