# Difference between revisions of "2005 AMC 8 Problems/Problem 17"

## Problem

The results of a cross-country team's training run are graphed below. Which student has the greatest average speed? $[asy] for ( int i = 1; i <= 7; ++i ) { draw((i,0)--(i,6)); } for ( int i = 1; i <= 5; ++i ) { draw((0,i)--(8,i)); } draw((-0.5,0)--(8,0), linewidth(1)); draw((0,-0.5)--(0,6), linewidth(1)); label("O", (0,0), SW); label(scale(.85)*rotate(90)*"distance", (0, 3), W); label(scale(.85)*"time", (4, 0), S); dot((1.25, 4.5)); label(scale(.85)*"Evelyn", (1.25, 4.8), N); dot((2.5, 2.2)); label(scale(.85)*"Briana", (2.5, 2.2), S); dot((4.25,5.2)); label(scale(.85)*"Carla", (4.25, 5.2), SE); dot((5.6, 2.8)); label(scale(.85)*"Debra", (5.6, 2.8), N); dot((6.8, 1.4)); label(scale(.85)*"Angela", (6.8, 1.4), E); [/asy]$

$\textbf{(A)}\ \text{Angela}\qquad\textbf{(B)}\ \text{Briana}\qquad\textbf{(C)}\ \text{Carla}\qquad\textbf{(D)}\ \text{Debra}\qquad\textbf{(E)}\ \text{Evelyn}$

## Solution

Average speed is distance over time, or the slope of the line through the point and the origin. $\boxed{\textbf{(E)}\ \text{Evelyn}}$ has the steepest line, and runs the greatest distance for the shortest amount of time.

 2005 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 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