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

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<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}}

Revision as of 20:53, 22 December 2012

Problem

The distance between the $5^\text{th}$ and $26^\text{th}$ exits on an interstate highway is $118$ miles. If any two 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 (ProblemsAnswer KeyResources)
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