2021 AMC 10A Problems/Problem 6

Problem

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 $4$ miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to $2$ miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at $3$ miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?

$\textbf{(A)} ~\frac{12}{13} \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~\frac{13}{12} \qquad\textbf{(D)} ~\frac{24}{13} \qquad\textbf{(E)} ~2$

Solution 1 (Generalized Distance)

Let $2d$ miles be the distance from the trailhead to the fire tower, where $d>0.$ When Chantal meets Jean, the two have traveled for $\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right) =d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d$ hours. At that point, Jean has traveled for $d$ miles, so his average speed is $\frac{d}{\frac{13}{12}d}=\boxed{\textbf{(A)} ~\frac{12}{13}}$ miles per hour.

~MRENTHUSIASM

Solution 2 (Specified Distance)

We will follow the same template as shown in Solution 1, except that we will replace $\boldsymbol{d}$ with a convenient constant.

Let $24$ miles be the distance from the trailhead to the fire tower. When Chantal meets Jean, the two have traveled for $\frac{12}{4} + \frac{12}{2}+\frac{12}{3}=3+6+4=13$ hours. At that point, Jean has traveled for $12$ miles, so his average speed is $\boxed{\textbf{(A)} ~\frac{12}{13}}$ miles per hour.

~MRENTHUSIASM

Solution 3 (d=st)

Shortened Solution

$d = st$ , so Jean's speed can be represented as $s = \frac{d}{t}$ . We know the time is $\left(\frac{d}{4} + \frac{d}{2} + \frac{d}{3}\right)$ (if the total distance is $2d$ ), so Jean's speed is $\frac{d}{(\frac{d}{4} + \frac{d}{2} + \frac{d}{3})} = \boxed{\textbf{(A)} ~\frac{12}{13}}$ .

Full Solution

We know that distance traveled is equal to the speed multiplied with the time. So, $d=st$ and $t = \frac{d}{s}$ . Let $2d$ be equal to the distance from the trailhead to the tower. Then, originally, Chantel travels a $d$ distance with at $4$ miles per hour. So, Chantel's time is $\frac{d}{4}$ . From the midpoint to the tower, Chantel takes $\frac{d}{2}$ hours (since Chantel has speed of $2$ miles per hour.) Similarly, the time it takes for Chantel to return to the midpoint is $\frac{d}{3}$ . Therefore, the total time is $\left(\frac{d}{4} + \frac{d}{2} + \frac{d}{3}\right)$ . We can can substitute this time into the original equation of $d=st$ to obtain $d = s\left(\frac{d}{4} + \frac{d}{2} + \frac{d}{3}\right)$ , so $\frac{d}{\left(\frac{d}{4} + \frac{d}{2} + \frac{d}{3}\right)} = s \implies \frac{1}{\frac{13}{12}} \implies \boxed{\textbf{(A)} ~\frac{12}{13}}$ .

~jaspersun

Video Solution 1 by OmegaLearn

https://youtu.be/hRFMsxhXQd0

~ pi_is_3.14

Video Solution 2 (Simple and Quick)

https://youtu.be/vwtGZVJ0TbI

~ Education, the Study of Everything

Video Solution 3

https://youtu.be/LonrTlNRk94

~savannahsolver

Video Solution 4 (by TheBeautyofMath)

https://youtu.be/cckGBU2x1zg

~IceMatrix

Video Solution by The Learning Royal

https://youtu.be/AWjOeBFyeb4

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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