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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:
  
 
<math> \textbf{(A)}\ b = g\qquad  
 
<math> \textbf{(A)}\ b = g\qquad  
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\textbf{(D)}\ b = g - 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 + g.}</math>
 
\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 ==
 
== Solution ==
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After inspection, we notice a general pattern: the <math>n^t^h</math> boy dances with <math>n + 4</math> girls.  
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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.
 
Since the last boy dances with all the girls, there must be four more girls than guys.
  
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== 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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