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 "1958 AHSME Problems/Problem 14"

m (→‎Solution)
m
Line 3: Line 3:
 
At a dance party a group of boys and girls exchange dances as follows: one boy dances with <math> 5</math> girls, a second boy dances with <math> 6</math> girls, and so on, the last boy dancing with all the girls. If <math> b</math> represents the number of boys and <math> g</math> the number of girls, then:
 
At a dance party a group of boys and girls exchange dances as follows: one boy dances with <math> 5</math> girls, a second boy dances with <math> 6</math> girls, and so on, the last boy dancing with all the girls. If <math> b</math> represents the number of boys and <math> g</math> the number of girls, then:
  
−
<math> \textbf{(A)}\ b \equal{} g\qquad  
+
<math> \textbf{(A)}\ b = g\qquad  
−
\textbf{(B)}\ b \equal{} \frac{g}{5}\qquad  
+
\textbf{(B)}\ b = \frac{g}{5}\qquad  
−
\textbf{(C)}\ b \equal{} g \minus{} 4\qquad  
+
\textbf{(C)}\ b = g - 4\qquad  
−
\textbf{(D)}\ b \equal{} g \minus{} 5\qquad \\
+
\textbf{(D)}\ b = g - 5\qquad \\
−
\textbf{(E)}\ \text{It is impossible to determine a relation between }{b}\text{ and }{g}\text{ without knowing }{b \plus{} g.}</math>
+
\textbf{(E)}\ \text{It is impossible to determine a relation between }{b}\text{ and }{g}\text{ without knowing }{b + g.}</math>
  
  

Revision as of 22:18, 13 March 2015

Problem

At a dance party a group of boys and girls exchange dances as follows: one boy dances with $5$ girls, a second boy dances with $6$ girls, and so on, the last boy dancing with all the girls. If $b$ represents the number of boys and $g$ the number of girls, then:

$\textbf{(A)}\ b = g\qquad \textbf{(B)}\ b = \frac{g}{5}\qquad \textbf{(C)}\ b = g - 4\qquad \textbf{(D)}\ b = g - 5\qquad \\ \textbf{(E)}\ \text{It is impossible to determine a relation between }{b}\text{ and }{g}\text{ without knowing }{b + g.}$


Solution

$\fbox{}$

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1958_AHSME_Problems/Problem_14&oldid=68947"