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> | ||
| − | ==Solution== | + | ==Solution 1== |
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. | 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== | ||
| + | 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. | ||
==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
Contents
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?
Solution 1
We can use the equation 
where 
is the distance, 
is the rate, and 
is the time. The distances he ran and walked are equal, so 
, where 
is the rate at which he ran, 
is the time for which he ran, 
is the rate at which he walked, and 
is the time for which he walked. Because he runs three times faster than he walks, 
. We want to find the time he ran, 
minutes. He traveled for a total of 
minutes.
Solution 2
We know that it took Joe 
minutes to walk halfway. If he ran 
times as fast as that, he went 
minutes for the other half (as 
and if we multiply 
by we are essentially dividing by ). Therefore, we have 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
