Difference between revisions of "2009 AMC 8 Problems/Problem 3"
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==Solution== | ==Solution== | ||
Suzanna's speed is <math>\frac{1}{5}</math>. This means she runs <math>\frac{1}{5} \cdot 30 = \boxed{ \textbf{(C) }6 }</math> | Suzanna's speed is <math>\frac{1}{5}</math>. This means she runs <math>\frac{1}{5} \cdot 30 = \boxed{ \textbf{(C) }6 }</math> | ||
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| + | ==Solution 2== | ||
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| + | From the graph, we can see that every <math>5</math> minutes Suzanna goes, her distance increases by <math>1</math>. Since half an hour is <math>10</math> minutes away, she would go <math>2</math> more miles. <math>4+2</math> is <math>6</math>, so the answer is C | ||
== Video Solution == | == Video Solution == | ||
Revision as of 20:46, 9 January 2023
Problem
The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
![[asy] import graph; /* this is a label */ Label f; f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0)); yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x) { return x; } draw(graph(f,-1,22),black+linewidth(1)); label("1", (-1,5), black); label("2", (-1, 10), black); label("3", (-1, 15), black); label("4", (-1, 20), black); dot((5,5), black+linewidth(5)); dot((10,10), black+linewidth(5)); dot((15, 15), black+linewidth(5)); dot((20,20), black+linewidth(5)); label("MINUTES", (11,-5), S); label(rotate(90)*"MILES", (-5,11), W);[/asy]](http://latex.artofproblemsolving.com/8/e/6/8e6a016e771b1190c29cf5d287fb3a89514557bc.png)
Solution
Suzanna's speed is 
. This means she runs 
Solution 2
From the graph, we can see that every 
minutes Suzanna goes, her distance increases by 
. Since half an hour is 
minutes away, she would go 
more miles. 
is 
, so the answer is C
Video Solution
https://youtu.be/USVVURBLaAc?t=117
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
