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 1 (Generalized Distance) ==
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==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)
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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)
=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.
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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
  
== Solution 2 (Convenient Distance) ==
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==Solution 2 (Specified Distance)==
<b>We use the same template as shown in Solution 1, except that we replace <math>\boldsymbol{d}</math> with a concrete number.</b>
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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 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.
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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 (Using Speed, Time, Distance) ==
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== Video Solution 1 (Using Speed, Time, Distance) ==
 
https://youtu.be/hRFMsxhXQd0
 
https://youtu.be/hRFMsxhXQd0
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution 2 (Simple and Quick)==
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https://youtu.be/vwtGZVJ0TbI
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 +
~ Education, the Study of Everything
  
 
==Video Solution 3==
 
==Video Solution 3==
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~savannahsolver
 
~savannahsolver
  
==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==
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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}}

Revision as of 20:31, 3 July 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 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

Video Solution 1 (Using Speed, Time, Distance)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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