Difference between revisions of "2021 AMC 10A Problems/Problem 6"
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~\frac{24}{13} \qquad\textbf{(E)} ~2</math> | ~\frac{24}{13} \qquad\textbf{(E)} ~2</math> | ||
| − | == Solution (Generalized Distance) == | + | == Solution 1 (Generalized Distance) == |
Let <math>2d</math> miles be the distance from the start to the fire tower. When Chantal meets Jean, she has traveled for <cmath>\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right) | Let <math>2d</math> miles be the distance from the start to the fire tower. When Chantal meets Jean, she has traveled for <cmath>\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right) | ||
=d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d</cmath> hours. Jean also has traveled for <math>\frac{13}{12}d</math> hours, and he travels for <math>d</math> miles. So, his average speed is <cmath>\frac{d}{\left(\frac{13}{12}d\right)}=\boxed{\textbf{(A)} ~\frac{12}{13}}</cmath> miles per hour. | =d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d</cmath> hours. Jean also has traveled for <math>\frac{13}{12}d</math> hours, and he travels for <math>d</math> miles. So, his average speed is <cmath>\frac{d}{\left(\frac{13}{12}d\right)}=\boxed{\textbf{(A)} ~\frac{12}{13}}</cmath> miles per hour. | ||
Revision as of 11:45, 12 February 2021
Contents
Problem 6
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at 
miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to 
miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at 
miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?
Solution 1 (Generalized Distance)
Let 
miles be the distance from the start to the fire tower. When Chantal meets Jean, she has traveled for ![\[\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right) =d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d\]](http://latex.artofproblemsolving.com/c/0/a/c0a9fe24b85132a945c1f2f286ed5334c47cc127.png)
hours. Jean also has traveled for 
hours, and he travels for 
miles. So, his average speed is ![\[\frac{d}{\left(\frac{13}{12}d\right)}=\boxed{\textbf{(A)} ~\frac{12}{13}}\]](http://latex.artofproblemsolving.com/a/b/4/ab474125fc3261d44530f8dc8043d32bc8aed757.png)
miles per hour.
~MRENTHUSIASM
Solution 2 (Convenient Distance)
Let 
miles be the distance from the start to the fire tower. When Chantal meets Jean, she travels for ![\[\frac{12}{4} + \frac{12}{2}+\frac{12}{3}=3+6+4=13\]](http://latex.artofproblemsolving.com/8/a/6/8a64535fec40239f4279ca8fb392c586b0977eaf.png)
hours. Jean also has traveled for 
hours, and he travels for miles. So, his average speed is 
miles per hour.
~MRENTHUSIASM
Video Solution (Using Speed, Time, Distance)
~ pi_is_3.14
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 AMC 10 Problems and Solutions | ||
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