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Difference between revisions of "2021 AMC 10A Problems/Problem 6"

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==Solution 1 (Generalized Distance)==
 
==Solution 1 (Generalized Distance)==
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Let <math>2d</math> miles be the distance from the trailhead to the fire tower. When Chantal meets Jean, the two have traveled for <cmath>\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right)
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Let <math>2d</math> miles be the distance from the trailhead to the fire tower, where <math>d>0.</math> When Chantal meets Jean, the two have traveled for <cmath>\frac d4 + \frac d2 + \frac d3 = d\left(\frac 14 + \frac 12 + \frac 13\right)
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=d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d</cmath> hours. Since Jean has traveled for <math>d</math> miles, his average speed is <math>\frac{d}{\left(\frac{13}{12}d\right)}=\boxed{\textbf{(A)} ~\frac{12}{13}}</math> miles per hour.
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=d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d</cmath> hours. At that point, Jean has traveled for <math>d</math> miles, so his average speed is <math>\frac{d}{\frac{13}{12}d}=\boxed{\textbf{(A)} ~\frac{12}{13}}</math> miles per hour.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2 (Convenient Distance)==
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==Solution 2 (Specified Distance)==
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<i><b>We will use the same template as shown in Solution 1, except that we will replace <math>\boldsymbol{d}</math> with a concrete number.</b></i>
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<i><b>We will follow the same template as shown in Solution 1, except that we will replace <math>\boldsymbol{d}</math> with a convenient constant.</b></i>
  
−
Let <math>24</math> miles be the distance from the trailhead to the fire tower. When Chantal meets Jean, she has traveled for <cmath>\frac{12}{4} + \frac{12}{2}+\frac{12}{3}=3+6+4=13</cmath> hours. Jean has also 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>24</math> miles be the distance from the trailhead to the fire tower. When Chantal meets Jean, the two have traveled for <cmath>\frac{12}{4} + \frac{12}{2}+\frac{12}{3}=3+6+4=13</cmath> hours. At that point, Jean has traveled for <math>12</math> miles, so his average speed is <math>\boxed{\textbf{(A)} ~\frac{12}{13}}</math> miles per hour.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Video Solution 2(Simple and Quick)==
+
==Solution 3 (d=st)==
−
https://youtu.be/vwtGZVJ0TbI
+
<i><b>Shortened Solution</b></I>
 +
 
 +
<math>d = st</math>, so Jean's speed can be represented as <math>s = \frac{d}{t}</math>. We know the time is <math>\left(\frac{d}{4} + \frac{d}{2} + \frac{d}{3}\right)</math> (if the total distance is <math>2d</math>), so Jean's speed is <math>\frac{d}{(\frac{d}{4} + \frac{d}{2} + \frac{d}{3})} = \boxed{\textbf{(A)} ~\frac{12}{13}}</math>.
 +
 
 +
<i><b>Full Solution</b></I>
  
−
~ Education, the Study of Everything
+
We know that distance traveled is equal to the speed multiplied with the time. So, <math>d=st</math> and <math>t = \frac{d}{s}</math>. Let <math>2d</math> be equal to the distance from the trailhead to the tower. Then, originally, Chantel travels a <math>d</math> distance with at <math>4</math> miles per hour. So, Chantel's time is <math>\frac{d}{4}</math>. From the midpoint to the tower, Chantel takes <math>\frac{d}{2}</math> hours (since Chantel has speed of <math>2</math> miles per hour.) Similarly, the time it takes for Chantel to return to the midpoint is <math>\frac{d}{3}</math>. Therefore, the total time is <math>\left(\frac{d}{4} + \frac{d}{2} + \frac{d}{3}\right)</math>. We can can substitute this time into the original equation of <math>d=st</math> to obtain <math>d = s\left(\frac{d}{4} + \frac{d}{2} + \frac{d}{3}\right)</math>, so <math>\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}}</math>.
  
 +
~jaspersun
  
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== Video Solution (Using Speed, Time, Distance) ==
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== Video Solution 1 by OmegaLearn ==
 
https://youtu.be/hRFMsxhXQd0
 
https://youtu.be/hRFMsxhXQd0
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution 2 (Simple and Quick)==
 +
https://youtu.be/vwtGZVJ0TbI
 +
 +
~ Education, the Study of Everything
  
 
==Video Solution 3==
 
==Video Solution 3==
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~savannahsolver
 
~savannahsolver
  
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==Video Solution by TheBeautyofMath==
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==Video Solution 4 (by TheBeautyofMath)==
 
https://youtu.be/cckGBU2x1zg
 
https://youtu.be/cckGBU2x1zg
  
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution by The Learning Royal==
 +
https://youtu.be/AWjOeBFyeb4
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2021|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:24, 3 September 2024

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

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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