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Difference between revisions of "2022 AMC 8 Problems/Problem 10"
Line 57: Line 57:
 
Note that:
 
Note that:
 
<ul style="list-style-type:square;">
 
<ul style="list-style-type:square;">
−
   <li>At <math>8 \, \textsc{am},</math> Ling's car was <math>0</math> miles from her house.</li><p>
+
   <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>
 
   <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>
−
It follows that at <math>10 \, \textsc{am},</math> Ling's car was <math>45\cdot2=90</math> miles from her house.</li><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>
 
   <li>From <math>10 \, \textsc{am}</math> to <math>1 \, \textsc{pm},</math> Ling did not move her car. <p>
−
It follows that at <math>1 \, \textsc{pm},</math> Ling's car was still <math>90</math> miles from her house.</li><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>
 
   <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>
−
It follows that at <math>2:30 \, \textsc{pm},</math> Ling's car was <math>0</math> miles from her house.</li><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>
 
</ul>
 
Therefore, the answer is <math>\boxed{\textbf{(E)}}.</math>
 
Therefore, the answer is <math>\boxed{\textbf{(E)}}.</math>

Revision as of 14:18, 28 January 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

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

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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