Difference between revisions of "2005 AMC 8 Problems/Problem 17"
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==Solution== | ==Solution== | ||
Average speed is distance over time, or the slope of the line through the point and the origin. <math>\boxed{\textbf{(E)}\ \text{Evelyn}}</math> has the steepest line, and runs the greatest distance for the shortest amount of time. | Average speed is distance over time, or the slope of the line through the point and the origin. <math>\boxed{\textbf{(E)}\ \text{Evelyn}}</math> has the steepest line, and runs the greatest distance for the shortest amount of time. | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=16|num-a=18}} | {{AMC8 box|year=2005|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:34, 24 September 2016
Problem
The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?
![[asy] for ( int i = 1; i <= 7; ++i ) { draw((i,0)--(i,6)); } for ( int i = 1; i <= 5; ++i ) { draw((0,i)--(8,i)); } draw((-0.5,0)--(8,0), linewidth(1)); draw((0,-0.5)--(0,6), linewidth(1)); label("$O$", (0,0), SW); label(scale(.85)*rotate(90)*"distance", (0, 3), W); label(scale(.85)*"time", (4, 0), S); dot((1.25, 4.5)); label(scale(.85)*"Evelyn", (1.25, 4.8), N); dot((2.5, 2.2)); label(scale(.85)*"Briana", (2.5, 2.2), S); dot((4.25,5.2)); label(scale(.85)*"Carla", (4.25, 5.2), SE); dot((5.6, 2.8)); label(scale(.85)*"Debra", (5.6, 2.8), N); dot((6.8, 1.4)); label(scale(.85)*"Angela", (6.8, 1.4), E); [/asy]](http://latex.artofproblemsolving.com/2/b/0/2b04d909f0586cae29f85ef276c64483f0e0c0e2.png)
Solution
Average speed is distance over time, or the slope of the line through the point and the origin. 
has the steepest line, and runs the greatest distance for the shortest amount of time.
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
