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

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==Problem==
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== Problem ==
 
The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?
 
The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?
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<asy>
 
<asy>
 
for ( int i = 1; i <= 7; ++i )
 
for ( int i = 1; i <= 7; ++i )
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<math> \textbf{(A)}\ \text{Angela}\qquad\textbf{(B)}\ \text{Briana}\qquad\textbf{(C)}\ \text{Carla}\qquad\textbf{(D)}\ \text{Debra}\qquad\textbf{(E)}\ \text{Evelyn} </math>
 
<math> \textbf{(A)}\ \text{Angela}\qquad\textbf{(B)}\ \text{Briana}\qquad\textbf{(C)}\ \text{Carla}\qquad\textbf{(D)}\ \text{Debra}\qquad\textbf{(E)}\ \text{Evelyn} </math>
  
==Solution==
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== 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.
  
==See Also==
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== See Also ==
 
{{AMC8 box|year=2005|num-b=16|num-a=18}}
 
{{AMC8 box|year=2005|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 12:34, 19 October 2020

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.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
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

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