Difference between revisions of "1992 AJHSME Problems/Problem 9"
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==Solution== | ==Solution== | ||
The graph show that the ratio of men to total population is <math>\frac{1}{3}</math>, so the total number of men is <math> \frac{1}{3} \times 480= \boxed{\text{(B)}\ 160} </math>. | The graph show that the ratio of men to total population is <math>\frac{1}{3}</math>, so the total number of men is <math> \frac{1}{3} \times 480= \boxed{\text{(B)}\ 160} </math>. | ||
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| + | ==Solution 2== | ||
| + | The graph shows <math>3</math> equal squares, each with value <math>x</math>. So <math>3x = 480</math>, so <math>x = \boxed{160}</math>. | ||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1992|num-b=8|num-a=10}} | {{AJHSME box|year=1992|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 06:08, 5 June 2020
Contents
Problem 9
The population of a small town is 
. The graph indicates the number of females and males in the town, but the vertical scale-values are omitted. How many males live in the town?
![[asy] draw((0,13)--(0,0)--(20,0)); draw((3,0)--(3,10)--(8,10)--(8,0)); draw((3,5)--(8,5)); draw((11,0)--(11,5)--(16,5)--(16,0)); label("$\textbf{POPULATION}$",(10,11),N); label("$\textbf{F}$",(5.5,0),S); label("$\textbf{M}$",(13.5,0),S); [/asy]](http://latex.artofproblemsolving.com/8/b/6/8b6893d40aebd402667ea599ed3ad0422b733c6b.png)
Solution
The graph show that the ratio of men to total population is 
, so the total number of men is 
.
Solution 2
The graph shows 
equal squares, each with value 
. So 
, so 
.
See Also
| 1992 AJHSME (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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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