2005 AMC 8 Problems/Problem 10

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.

2005 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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2005 AMC 8 Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

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