Difference between revisions of "1992 AJHSME Problems/Problem 18"
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==Solution== | ==Solution== | ||
The average speed is given by the total distance traveled divided by the total time traveled. | The average speed is given by the total distance traveled divided by the total time traveled. | ||
| − | <cmath>\frac{80+0+100}{4} = \boxed{\text{(A)}\ 45}</cmath> | + | <cmath>\frac{80+0+100}{4} = \boxed{\text{(\textbf{A})}\ 45}</cmath> |
==See Also== | ==See Also== | ||
{{AJHSME box|year=1992|num-b=17|num-a=19}} | {{AJHSME box|year=1992|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 04:42, 31 August 2015
Problem
On a trip, a car traveled 
miles in an hour and a half, then was stopped in traffic for 
minutes, then traveled 
miles during the next 
hours. What was the car's average speed in miles per hour for the 
-hour trip?
Solution
The average speed is given by the total distance traveled divided by the total time traveled.
![\[\frac{80+0+100}{4} = \boxed{\text{(\textbf{A})}\ 45}\]](http://latex.artofproblemsolving.com/d/e/f/def274faa9158edbd089074013c158a2bf1b273a.png)
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
