Difference between revisions of "1983 AHSME Problems/Problem 9"
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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>. | 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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1983|num-b=8|num-a=10}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 23:45, 19 February 2019
Problem
In a certain population the ratio of the number of women to the number of men is 
to 
.
If the average (arithmetic mean) age of the women is 
and the average age of the men is 
,
then the average age of the population is
Solution
Assume, without loss of generality, that there are exactly women and men. Then the total age of the women is 
and the total age of the men is 
. Therefore the overall average is 
.
See Also
| 1983 AHSME (Problems • Answer Key • Resources) | ||
| 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. 
