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

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== Problem ==
 
== Problem ==
  
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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:
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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:
  
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<math> \textbf{(A)}\ b \equal{} g\qquad  
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<math> \textbf{(A)}\ b = g\qquad  
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\textbf{(B)}\ b \equal{} \frac{g}{5}\qquad  
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\textbf{(B)}\ b = \frac{g}{5}\qquad  
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\textbf{(C)}\ b \equal{} g \minus{} 4\qquad  
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\textbf{(C)}\ b = g - 4\qquad  
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\textbf{(D)}\ b \equal{} g \minus{} 5\qquad \\
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\textbf{(D)}\ b = g - 5\qquad \\
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\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>
  
 +
== 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.
  
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== Solution ==
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Therefore, the equation that relates them is <math>\fbox{(C) b = g - 4}</math>
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<math>\fbox{}</math>
 
  
 
== See Also ==
 
== See Also ==
  
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{{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 20:25, 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 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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

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