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

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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  
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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>
  
  

Revision as of 23:18, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AHSME Problems and Solutions

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