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Difference between revisions of "2021 AMC 10A Problems/Problem 6"
(→‎Solution 2 (Convenient Distance))
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== Solution 2 (Convenient Distance) ==
 
== Solution 2 (Convenient Distance) ==
 +
We use the same template as shown in Solution 1, except that we replace <math>d</math> with a concrete number.
 +
 
Let <math>12</math> miles be the distance from the start to the fire tower. When Chantal meets Jean, she travels for <cmath>\frac{12}{4} + \frac{12}{2}+\frac{12}{3}=3+6+4=13</cmath> hours. Jean also has traveled for <math>13</math> hours, and he travels for <math>12</math> miles. So, his average speed is <math>\boxed{\textbf{(A)} ~\frac{12}{13}}</math> miles per hour.
 
Let <math>12</math> miles be the distance from the start to the fire tower. When Chantal meets Jean, she travels for <cmath>\frac{12}{4} + \frac{12}{2}+\frac{12}{3}=3+6+4=13</cmath> hours. Jean also has traveled for <math>13</math> hours, and he travels for <math>12</math> miles. So, his average speed is <math>\boxed{\textbf{(A)} ~\frac{12}{13}}</math> miles per hour.
  

Revision as of 18:06, 12 February 2021

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 start to the fire tower. When Chantal meets Jean, she has traveled for \[\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right) =d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d\] hours. Jean also has traveled for $\frac{13}{12}d$ hours, and he travels for $d$ miles. So, his average speed is \[\frac{d}{\left(\frac{13}{12}d\right)}=\boxed{\textbf{(A)} ~\frac{12}{13}}\] miles per hour.

~MRENTHUSIASM

Solution 2 (Convenient Distance)

We use the same template as shown in Solution 1, except that we replace $d$ with a concrete number.

Let $12$ miles be the distance from the start to the fire tower. When Chantal meets Jean, she travels for \[\frac{12}{4} + \frac{12}{2}+\frac{12}{3}=3+6+4=13\] hours. Jean also has traveled for $13$ hours, and he travels for $12$ miles. So, his average speed is $\boxed{\textbf{(A)} ~\frac{12}{13}}$ miles per hour.

~MRENTHUSIASM

Video Solution (Using Speed, Time, Distance)

https://youtu.be/hRFMsxhXQd0

~ 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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