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

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One sunny day, Ling decided to take a hike in the mountains. She left her house at <math>8 \, \textsc{am}</math>, drove at a constant speed of <math>45</math> miles per hour, and arrived at the hiking trail at <math>10 \, \textsc{am}</math>. After hiking for <math>3</math> hours, Ling drove home at a constant speed of <math>60</math> 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?
 
One sunny day, Ling decided to take a hike in the mountains. She left her house at <math>8 \, \textsc{am}</math>, drove at a constant speed of <math>45</math> miles per hour, and arrived at the hiking trail at <math>10 \, \textsc{am}</math>. After hiking for <math>3</math> hours, Ling drove home at a constant speed of <math>60</math> 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?
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+
 
 
<asy>
 
<asy>
 
unitsize(12);
 
unitsize(12);
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}
 
}
 
draw(rotate(90)*"Distance (miles)",sh+(-2.1,3),fontsize(10));
 
draw(rotate(90)*"Distance (miles)",sh+(-2.1,3),fontsize(10));
−
label("$\textbf{("+lab+")}$",sh+(-2.1,6.8),fontsize(10));
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label("$\textbf{("+lab+")}$",sh+(-2.1,6.8),fontsize(12));
 
}
 
}
 
drawgraph((0,0),"A");
 
drawgraph((0,0),"A");
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</asy>
 
</asy>
  
−
==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>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/Ij9pAy6tQSg?t=733
 +
 
 +
~Interstigation
  
 
==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}}

Revision as of 00:16, 8 April 2022

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 $\boxed{\textbf{(E)}}$.

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=733

~Interstigation

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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