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 "2004 AMC 8 Problems/Problem 20"

(→‎Solution)
Line 6: Line 6:
 
==Solution 1==
 
==Solution 1==
 
Working backwards, if <math>3/4</math> of the chairs are taken and <math>6</math> are empty, then there are three times as many taken chairs as empty chairs, or <math>3 \cdot 6 = 18</math>. If <math>x</math> is the number of people in the room and <math>2/3</math> are seated, then <math>\frac23 x = 18</math> and <math>x = \boxed{(\text{D}) 27}</math>.
 
Working backwards, if <math>3/4</math> of the chairs are taken and <math>6</math> are empty, then there are three times as many taken chairs as empty chairs, or <math>3 \cdot 6 = 18</math>. If <math>x</math> is the number of people in the room and <math>2/3</math> are seated, then <math>\frac23 x = 18</math> and <math>x = \boxed{(\text{D}) 27}</math>.
 +
 +
==Video Solution==
 +
https://youtu.be/GhrphDrsvl4 Soo, DRMS, NM
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2004|num-b=19|num-a=21}}
 
{{AMC8 box|year=2004|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:43, 25 March 2022

Problem

Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution 1

Working backwards, if $3/4$ of the chairs are taken and $6$ are empty, then there are three times as many taken chairs as empty chairs, or $3 \cdot 6 = 18$. If $x$ is the number of people in the room and $2/3$ are seated, then $\frac23 x = 18$ and $x = \boxed{(\text{D}) 27}$.

Video Solution

https://youtu.be/GhrphDrsvl4 Soo, DRMS, NM

See Also

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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=2004_AMC_8_Problems/Problem_20&oldid=172931"