Difference between revisions of "1958 AHSME Problems/Problem 14"

m (Solution)
m (Problem)
 
(5 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
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: The first 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>
  
 +
== Solution ==
 +
After inspection, we notice a general pattern: the <math>nth</math> boy dances with <math>n + 4</math> girls.
 +
Since the last boy dances with all the girls, there must be four more girls than guys.
  
== Solution ==
+
Therefore, the equation that relates them is <math>\fbox{(C) b = g - 4}</math>
<math>\fbox{}</math>
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 23:04, 19 February 2020

Problem

At a dance party a group of boys and girls exchange dances as follows: The first 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

After inspection, we notice a general pattern: the $nth$ boy dances with $n + 4$ girls. Since the last boy dances with all the girls, there must be four more girls than guys.

Therefore, the equation that relates them is $\fbox{(C) b = g - 4}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS