Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1992 AJHSME Problems/Problem 9"
m (→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 19: Line 19:
 
==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>.
 +
 +
==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}}

Latest revision as of 05:08, 5 June 2020

Problem 9

The population of a small town is $480$. 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]

$\text{(A)}\ 120 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 360$

Solution

The graph show that the ratio of men to total population is $\frac{1}{3}$, so the total number of men is $\frac{1}{3} \times 480= \boxed{\text{(B)}\ 160}$.

Solution 2

The graph shows $3$ equal squares, each with value $x$. So $3x = 480$, so $x = \boxed{160}$.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AJHSME_Problems/Problem_9&oldid=123867"