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