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Difference between revisions of "2022 AMC 8 Problems/Problem 10"

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</asy>
  
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==Solution==
+
==Solution 1 (Analysis)==
 +
 
 +
Note that:
 +
<ul style="list-style-type:square;">
 +
  <li><b>At <math>\boldsymbol{8 \, \footnotesize\textbf{AM},}</math> Ling's car was <math>\boldsymbol{0}</math> miles from her house.</b></li><p>
 +
  <li>From <math>8 \, \textsc{am}</math> to <math>10 \, \textsc{am},</math> Ling drove to the hiking trail at a constant speed of <math>45</math> miles per hour. <p>
 +
<b>It follows that at <math>\boldsymbol{10 \, \footnotesize\textbf{AM},}</math> Ling's car was <math>\boldsymbol{45\cdot2=90}</math> miles from her house.</b></li><p>
 +
  <li>From <math>10 \, \textsc{am}</math> to <math>1 \, \textsc{pm},</math> Ling did not move her car. <p>
 +
<b>It follows that at <math>\boldsymbol{1 \, \footnotesize\textbf{PM},}</math> Ling's car was still <math>\boldsymbol{90}</math> miles from her house.</b></li><p>
 +
  <li>From <math>1 \, \textsc{pm},</math> Ling drove home at a constant speed of <math>60</math> miles per hour. So, she arrived home <math>90\div60=1.5</math> hour later. <p>
 +
<b>It follows that at <math>\boldsymbol{2:30 \, \footnotesize\textbf{PM},}</math> Ling's car was <math>\boldsymbol{0}</math> miles from her house.</b></li><p>
 +
</ul>
 +
Therefore, the answer is <math>\boxed{\textbf{(E)}}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Solution 2 (Elimination)==
 +
Ling's trip took <math>2</math> hours, thus she traveled for <math>2 \times 45=90</math> miles. Choices <math>\textbf{(B)}</math>, <math>\textbf{(C)}</math>, and <math>\textbf{(D)}</math> are eliminated.
 +
Ling drove <math>45</math> miles per hour (mph) to the mountains, and <math>60</math> mph back to her house, so the rightmost slope must be steeper than the leftmost one. Choice <math>\textbf {(A)}</math> is eliminated.
 +
This leaves us with <math>\boxed{\textbf{(E)}}</math>.
 +
 
 +
==Solution 3 (Elimination)==
 +
 
 +
Using the <math>\text{speed} = \frac{\text{distance}}{\text{time}}</math> formula, and plugging in the values <math>45</math> mph and <math>2</math> hrs, we get that the distance from Ling's house to the mountains is <math>90</math> miles. That means the first slope ends at <math>90,</math> so choices <math>\textbf{(B)}, \textbf{(C)},</math> and <math>\textbf{(D)}</math> are eliminated.
 +
 
 +
Furthermore, if the distance is <math>90</math> miles, and Ling is returning home at <math>60</math> mph, it must have taken her <math>1.5</math> hours. Adding <math>1.5</math> and <math>3</math> (how long she hiked for) to her arrival time, <math>10</math> AM, we see she must have come back home at <math>2:30</math> PM. Choice <math>\textbf{(A)}</math> is eliminated, so the only valid choice left is choice <math>\boxed{\textbf{(E)}}.</math>
 +
 
 +
~ProProtractor
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/oUEa7AjMF2A?si=GriQR9m9WenwQVdf&t=1311
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (CRITICAL THINKING!!!)==
 +
https://youtu.be/Q3G-qyCUnYI
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=733
 +
 
 +
~Interstigation
 +
 
 +
==Video Solution==
 +
https://youtu.be/1xspUFoKDnU?t=332
 +
 
 +
~STEMbreezy
 +
 
 +
==Video Solution==
 +
https://youtu.be/q4hJk1HYxDk
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
https://youtu.be/BzKZSwxJHJ0
 +
 
 +
~harungurcan
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=9|num-a=11}}
 
{{AMC8 box|year=2022|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 03:59, 22 January 2024

Problem

One sunny day, Ling decided to take a hike in the mountains. She left her house at $8 \, \textsc{am}$, drove at a constant speed of $45$ miles per hour, and arrived at the hiking trail at $10 \, \textsc{am}$. After hiking for $3$ hours, Ling drove home at a constant speed of $60$ miles per hour. Which of the following graphs best illustrates the distance between Ling’s car and her house over the course of her trip?

[asy] unitsize(12); usepackage("mathptmx"); defaultpen(fontsize(8)+linewidth(.7)); int mod12(int i) {if (i<13) {return i;} else {return i-12;}} void drawgraph(pair sh,string lab) { for (int i=0;i<11;++i) { for (int j=0;j<6;++j) { draw(shift(sh+(i,j))*unitsquare,mediumgray); } } draw(shift(sh)*((-1,0)--(11,0)),EndArrow(angle=20,size=8)); draw(shift(sh)*((0,-1)--(0,6)),EndArrow(angle=20,size=8)); for (int i=1;i<10;++i) { draw(shift(sh)*((i,-.2)--(i,.2))); } label("8\tiny{\textsc{am}}",sh+(1,-.2),S); for (int i=2;i<9;++i) { label(string(mod12(i+7)),sh+(i,-.2),S); } label("4\tiny{\textsc{pm}}",sh+(9,-.2),S); for (int i=1;i<6;++i) { label(string(30*i),sh+(0,i),2*W); } draw(rotate(90)*"Distance (miles)",sh+(-2.1,3),fontsize(10)); label("$\textbf{("+lab+")}$",sh+(-2.1,6.8),fontsize(12)); } drawgraph((0,0),"A"); drawgraph((15,0),"B"); drawgraph((0,-10),"C"); drawgraph((15,-10),"D"); drawgraph((0,-20),"E"); dotfactor=6; draw((1,0)--(3,3)--(6,3)--(8,0),linewidth(.9)); dot((1,0)^^(3,3)^^(6,3)^^(8,0)); pair sh = (15,0); draw(shift(sh)*((1,0)--(3,1.5)--(6,1.5)--(8,0)),linewidth(.9)); dot(sh+(1,0)^^sh+(3,1.5)^^sh+(6,1.5)^^sh+(8,0)); pair sh = (0,-10); draw(shift(sh)*((1,0)--(3,1.5)--(6,1.5)--(7.5,0)),linewidth(.9)); dot(sh+(1,0)^^sh+(3,1.5)^^sh+(6,1.5)^^sh+(7.5,0)); pair sh = (15,-10); draw(shift(sh)*((1,0)--(3,4)--(6,4)--(9.3,0)),linewidth(.9)); dot(sh+(1,0)^^sh+(3,4)^^sh+(6,4)^^sh+(9.3,0)); pair sh = (0,-20); draw(shift(sh)*((1,0)--(3,3)--(6,3)--(7.5,0)),linewidth(.9)); dot(sh+(1,0)^^sh+(3,3)^^sh+(6,3)^^sh+(7.5,0)); [/asy]

Solution 1 (Analysis)

Note that:

  • At $\boldsymbol{8 \, \footnotesize\textbf{AM},}$ Ling's car was $\boldsymbol{0}$ miles from her house.

  • From $8 \, \textsc{am}$ to $10 \, \textsc{am},$ Ling drove to the hiking trail at a constant speed of $45$ miles per hour.

    It follows that at $\boldsymbol{10 \, \footnotesize\textbf{AM},}$ Ling's car was $\boldsymbol{45\cdot2=90}$ miles from her house.

  • From $10 \, \textsc{am}$ to $1 \, \textsc{pm},$ Ling did not move her car.

    It follows that at $\boldsymbol{1 \, \footnotesize\textbf{PM},}$ Ling's car was still $\boldsymbol{90}$ miles from her house.

  • From $1 \, \textsc{pm},$ Ling drove home at a constant speed of $60$ miles per hour. So, she arrived home $90\div60=1.5$ hour later.

    It follows that at $\boldsymbol{2:30 \, \footnotesize\textbf{PM},}$ Ling's car was $\boldsymbol{0}$ miles from her house.

Therefore, the answer is $\boxed{\textbf{(E)}}.$

~MRENTHUSIASM

Solution 2 (Elimination)

Ling's trip took $2$ hours, thus she traveled for $2 \times 45=90$ miles. Choices $\textbf{(B)}$, $\textbf{(C)}$, and $\textbf{(D)}$ are eliminated. Ling drove $45$ miles per hour (mph) to the mountains, and $60$ mph back to her house, so the rightmost slope must be steeper than the leftmost one. Choice $\textbf {(A)}$ is eliminated. This leaves us with $\boxed{\textbf{(E)}}$.

Solution 3 (Elimination)

Using the $\text{speed} = \frac{\text{distance}}{\text{time}}$ formula, and plugging in the values $45$ mph and $2$ hrs, we get that the distance from Ling's house to the mountains is $90$ miles. That means the first slope ends at $90,$ so choices $\textbf{(B)}, \textbf{(C)},$ and $\textbf{(D)}$ are eliminated.

Furthermore, if the distance is $90$ miles, and Ling is returning home at $60$ mph, it must have taken her $1.5$ hours. Adding $1.5$ and $3$ (how long she hiked for) to her arrival time, $10$ AM, we see she must have come back home at $2:30$ PM. Choice $\textbf{(A)}$ is eliminated, so the only valid choice left is choice $\boxed{\textbf{(E)}}.$

~ProProtractor

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=GriQR9m9WenwQVdf&t=1311

~Math-X

Video Solution (CRITICAL THINKING!!!)

https://youtu.be/Q3G-qyCUnYI

~Education, the Study of Everything


Video Solution

https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=733

~Interstigation

Video Solution

https://youtu.be/1xspUFoKDnU?t=332

~STEMbreezy

Video Solution

https://youtu.be/q4hJk1HYxDk

~savannahsolver

Video Solution

https://youtu.be/BzKZSwxJHJ0

~harungurcan

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 AJHSME/AMC 8 Problems and Solutions

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