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

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== 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:
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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  
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<math> \textbf{(A)}\ b = g\qquad  
\textbf{(B)}\ b \equal{} \frac{g}{5}\qquad  
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\textbf{(B)}\ b = \frac{g}{5}\qquad  
\textbf{(C)}\ b \equal{} g \minus{} 4\qquad  
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\textbf{(C)}\ b = g - 4\qquad  
\textbf{(D)}\ b \equal{} g \minus{} 5\qquad \\
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\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>
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\textbf{(E)}\ \text{It is impossible to determine a relation between }{b}\text{ and }{g}\text{ without knowing }{b + g.}</math>
  
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== Solution ==
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After inspection, we notice a general pattern: the <math>nth</math> boy dances with <math>n + 4</math> girls.
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Since the last boy dances with all the girls, there must be four more girls than guys.
  
== Solution ==
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Therefore, the equation that relates them is <math>\fbox{(C) b = g - 4}</math>
\fbox{}
 
  
 
== See Also ==
 
== See Also ==
  
{{AHSME 50p box|year=1958|num-b=13|num-a=15}}
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{{AHSME 50p box|year=1958|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:25, 27 January 2023

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 14
Followed by
Problem 16
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All AHSME Problems and Solutions

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