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Difference between revisions of "2005 AMC 8 Problems/Problem 10"
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<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3 </math>
 
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3 </math>
  
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==Solution==
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==Solution 1==
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Use the equation <math>d=rt</math> where <math>d</math> is the distance, <math>r</math> is the rate, and <math>t</math> is the time. The distances he ran and walked are equal, so <math>r_rt_r=r_wt_w</math>. Because he runs three times faster than he walks, <math>r_r=3r_w</math>. We want to find the time he ran, <math>t_r = \frac{r_wt_w}{t_r} = \frac{(r_w)(6)}{3r_w} = 2</math> minutes. He traveled for a total of <math>6+2=\boxed{\textbf{(D)}\ 8}</math> minutes.
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We can use the equation <math>d=rt</math> where <math>d</math> is the distance, <math>r</math> is the rate, and <math>t</math> is the time. The distances he ran and walked are equal, so <math>r_rt_r=r_wt_w</math>, where <math>r_r</math> is the rate at which he ran, <math>t_r</math> is the time for which he ran, <math>r_w</math> is the rate at which he walked, and <math>t_w</math> is the time for which he walked. Because he runs three times faster than he walks, <math>r_r=3r_w</math>. We want to find the time he ran, <math>t_r = \frac{r_wt_w}{t_r} = \frac{(r_w)(6)}{3r_w} = 2</math> minutes. He traveled for a total of <math>6+2=\boxed{\textbf{(D)}\ 8}</math> minutes.
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==Solution 2==
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We know that it took Joe <math>6</math> minutes to walk halfway. If he ran <math>3</math> times as fast as that, he went <math>\frac63 = 2</math> minutes for the other half (as <math>t=\frac{d}{s}</math> and if we multiply <math>s</math> by <math>3</math> we are essentially dividing <math>t</math> by <math>3</math>). Therefore, we have <math>6 + 2 = 8</math> minutes.
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==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=9|num-a=11}}
 
{{AMC8 box|year=2005|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:29, 23 July 2024

Problem

Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3$

Solution 1

We can use the equation $d=rt$ where $d$ is the distance, $r$ is the rate, and $t$ is the time. The distances he ran and walked are equal, so $r_rt_r=r_wt_w$, where $r_r$ is the rate at which he ran, $t_r$ is the time for which he ran, $r_w$ is the rate at which he walked, and $t_w$ is the time for which he walked. Because he runs three times faster than he walks, $r_r=3r_w$. We want to find the time he ran, $t_r = \frac{r_wt_w}{t_r} = \frac{(r_w)(6)}{3r_w} = 2$ minutes. He traveled for a total of $6+2=\boxed{\textbf{(D)}\ 8}$ minutes.

Solution 2

We know that it took Joe $6$ minutes to walk halfway. If he ran $3$ times as fast as that, he went $\frac63 = 2$ minutes for the other half (as $t=\frac{d}{s}$ and if we multiply $s$ by $3$ we are essentially dividing $t$ by $3$). Therefore, we have $6 + 2 = 8$ minutes.

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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