Difference between revisions of "1992 AJHSME Problems/Problem 18"

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<math>\text{(A)}\ 45 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90</math>
 
<math>\text{(A)}\ 45 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90</math>
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==Solution==
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The average speed is given by the total distance traveled divided by the total time traveled.
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<cmath>\frac{80+0+100}{4} = \boxed{\text{(\textbf{A})}\ 45}</cmath>
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==See Also==
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{{AJHSME box|year=1992|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 05:42, 31 August 2015

Problem

On a trip, a car traveled $80$ miles in an hour and a half, then was stopped in traffic for $30$ minutes, then traveled $100$ miles during the next $2$ hours. What was the car's average speed in miles per hour for the $4$-hour trip?

$\text{(A)}\ 45 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90$

Solution

The average speed is given by the total distance traveled divided by the total time traveled. \[\frac{80+0+100}{4} = \boxed{\text{(\textbf{A})}\ 45}\]

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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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