# 2022 AMC 8 Problems/Problem 10

## 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

~Interstigation

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