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Difference between revisions of "2009 AMC 8 Problems/Problem 3"

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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 <math>\boxed{ \textbf{(C) }6 }</math>
 
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 <math>\boxed{ \textbf{(C) }6 }</math>
 +
 +
~Trex226
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 16:18, 10 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]

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7$

Solution

Suzanna's speed is $\frac{1}{5}$. This means she runs $\frac{1}{5} \cdot 30 = \boxed{ \textbf{(C) }6 }$

Solution 2

From the graph, we can see that every $5$ minutes Suzanna goes, her distance increases by $1$. Since half an hour is $10$ minutes away, she would go $2$ more miles. $4+2$ is $6$, so the answer is $\boxed{ \textbf{(C) }6 }$

~Trex226

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

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