Difference between revisions of "2021 AMC 10A Problems/Problem 6"
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Convenient Distance): Bolded the variable.) |
MRENTHUSIASM (talk | contribs) (→Solution 3 (d=st)) |
||
(33 intermediate revisions by 7 users not shown) | |||
Line 5: | Line 5: | ||
~\frac{24}{13} \qquad\textbf{(E)} ~2</math> | ~\frac{24}{13} \qquad\textbf{(E)} ~2</math> | ||
− | == Solution 1 (Generalized Distance) == | + | ==Solution 1 (Generalized Distance)== |
− | Let <math>2d</math> miles be the distance from the | + | 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 | + | =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 ( | + | ==Solution 2 (Specified Distance)== |
− | <b>We | + | <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 | + | 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 | ||
− | == | + | ==Solution 3 (d=st)== |
+ | <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> | ||
+ | |||
+ | 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>3</math> miles per hour. So, Chantel's time is <math>\frac{d}{3}</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 | ||
+ | |||
+ | == 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)== | + | ==Video Solution 2 (Simple and Quick)== |
https://youtu.be/vwtGZVJ0TbI | https://youtu.be/vwtGZVJ0TbI | ||
Line 33: | Line 44: | ||
~savannahsolver | ~savannahsolver | ||
− | ==Video Solution by TheBeautyofMath== | + | ==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 03:34, 19 October 2023
Contents
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 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 trailhead to the fire tower, where
When Chantal meets Jean, the two have traveled for
hours. At that point, Jean has traveled for
miles, so his average speed is
miles per hour.
~MRENTHUSIASM
Solution 2 (Specified Distance)
We will follow the same template as shown in Solution 1, except that we will replace with a convenient constant.
Let miles be the distance from the trailhead to the fire tower. When Chantal meets Jean, the two have traveled for
hours. At that point, Jean has traveled for
miles, so his average speed is
miles per hour.
~MRENTHUSIASM
Solution 3 (d=st)
Shortened Solution
, so Jean's speed can be represented as
. We know the time is
(if the total distance is
), so Jean's speed is
.
Full Solution
We know that distance traveled is equal to the speed multiplied with the time. So, and
. Let
be equal to the distance from the trailhead to the tower. Then, originally, Chantel travels a
distance with at
miles per hour. So, Chantel's time is
. From the midpoint to the tower, Chantel takes
hours (since Chantel has speed of
miles per hour.) Similarly, the time it takes for Chantel to return to the midpoint is
. Therefore, the total time is
. We can can substitute this time into the original equation of
to obtain
, so
.
~jaspersun
Video Solution 1 by OmegaLearn
~ pi_is_3.14
Video Solution 2 (Simple and Quick)
~ Education, the Study of Everything
Video Solution 3
~savannahsolver
Video Solution 4 (by TheBeautyofMath)
~IceMatrix
Video Solution by The Learning Royal
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.