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

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

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

Latest revision as of 03:59, 22 January 2024

Contents

Problem

Solution 1 (Analysis)

Solution 2 (Elimination)

Solution 3 (Elimination)

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

Video Solution (CRITICAL THINKING!!!)

Video Solution

Video Solution

Video Solution

Video Solution

See Also

@@ Line 1: / Line 1: @@
 ==Problem==
-COPY AND PASTE THE PROBLEM HERE.
-==Solution==
+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?
+<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:
+<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==
 {{AMC8 box|year=2022|num-b=9|num-a=11}}
 {{MAA Notice}}