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Difference between revisions of "1992 AJHSME Problems/Problem 19"

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==Problem==
 
==Problem==
  
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The distance between the <math>5^\text{th}</math> and <math>26^\text{th}</math> exits on an interstate highway is <math>118</math> miles.  If any two exits are at least <math>5</math> miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the <math>5^\text{th}</math> and <math>26^\text{th}</math> exits?
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The distance between the <math>5^\text{th}</math> and <math>26^\text{th}</math> exits on an interstate highway is <math>118</math> miles.  If any two consecutive exits are at least <math>5</math> miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the <math>5^\text{th}</math> and <math>26^\text{th}</math> exits?
  
 
<math>\text{(A)}\ 8 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 47 \qquad \text{(E)}\ 98</math>
 
<math>\text{(A)}\ 8 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 47 \qquad \text{(E)}\ 98</math>
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==Solution==
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There are <math>21</math> pairs of consecutive exits. To find the maximum number of miles of one of these, the other <math>20</math> must be equal to the minimum number yielding a total of <math>(5)(20)=100</math> miles. The longest distance must be <math>118-100=\boxed{\text{(C)}\ 18}</math>.
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==See Also==
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{{AJHSME box|year=1992|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 21:26, 25 May 2021

Problem

The distance between the $5^\text{th}$ and $26^\text{th}$ exits on an interstate highway is $118$ miles. If any two consecutive exits are at least $5$ miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the $5^\text{th}$ and $26^\text{th}$ exits?

$\text{(A)}\ 8 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 47 \qquad \text{(E)}\ 98$

Solution

There are $21$ pairs of consecutive exits. To find the maximum number of miles of one of these, the other $20$ must be equal to the minimum number yielding a total of $(5)(20)=100$ miles. The longest distance must be $118-100=\boxed{\text{(C)}\ 18}$.

See Also

1992 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
All AJHSME/AMC 8 Problems and Solutions

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