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Difference between revisions of "2023 AMC 8 Problems/Problem 13"
(→‎Video Solution by Math-X (Let's first Understand the question))
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==Problem==
  
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Along the route of a bicycle race, <math>7</math> water stations are evenly spaced between the start and finish lines,
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as shown in the figure below. There are also <math>2</math> repair stations evenly spaced between the start and
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finish lines. The <math>3</math>rd water station is located <math>2</math> miles after the <math>1</math>st repair station. How long is the race
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in miles?
 +
<asy>
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// Credits given to Themathguyd‎ and Kante314
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usepackage("mathptmx");
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size(10cm);
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filldraw((11,4.5)--(171,4.5)--(171,17.5)--(11,17.5)--cycle,mediumgray*0.4 + lightgray*0.6);
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draw((11,11)--(171,11),linetype("2 2")+white+linewidth(1.2));
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draw((0,0)--(11,0)--(11,22)--(0,22)--cycle);
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draw((171,0)--(182,0)--(182,22)--(171,22)--cycle);
  
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==Written Solution==
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draw((31,4.5)--(31,0));
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draw((51,4.5)--(51,0));
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draw((151,4.5)--(151,0));
  
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Knowing that there are <math>7</math> equally spaced water stations they are each located <math>\frac{d}{8}</math>, <math>\frac{2d}{8}</math>,… <math>\frac{7d}{8}</math> of the way from the start. Using the same logic for the <math>3</math> station we have <math>\frac{d}{3}</math> and <math>\frac{2d}{3}</math> for the repair stations. It is given that the 3rd water is <math>2</math> miles ahead of the <math>1</math>st repair station. So setting an equation we have <math>\frac{3d}{8} = \frac{d}{3} + 2</math> getting common denominators <math>\frac{9d}{24} = \frac{8d}{24} + 2</math> so then we have <math>d =  \boxed{\text{(D)}48}</math> from this.
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label(scale(.85)*rotate(45)*"Water 1", (23,-13.5));
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label(scale(.85)*rotate(45)*"Water 2", (43,-13.5));
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label(scale(.85)*rotate(45)*"Water 7", (143,-13.5));
  
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
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filldraw(circle((103,-13.5),.2));
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filldraw(circle((98,-13.5),.2));
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filldraw(circle((93,-13.5),.2));
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filldraw(circle((88,-13.5),.2));
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filldraw(circle((83,-13.5),.2));
  
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label(scale(.85)*rotate(90)*"Start", (5.5,11));
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label(scale(.85)*rotate(270)*"Finish", (176.5,11));
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</asy>
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<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 96</math>
  
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==Animated Video Solution==
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==Solution==
 +
 
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Suppose that the race is <math>d</math> miles long. The water stations are located at <cmath>\frac{d}{8}, \frac{2d}{8}, \ldots, \frac{7d}{8}</cmath> miles from the start, and the repair stations are located at <cmath>\frac{d}{3}, \frac{2d}{3}</cmath> miles from the start.
 +
 
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We are given that <math>\frac{3d}{8}=\frac{d}{3}+2,</math> from which 
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<cmath>\begin{align*}
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\frac{9d}{24}&=\frac{8d}{24}+2 \\
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\frac{d}{24}&=2 \\
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d&=\boxed{\textbf{(D)}\ 48}.
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\end{align*}</cmath>
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
 +
 
 +
==Video Solution by Math-X (Let's first Understand the question)==
 +
https://youtu.be/Ku_c1YHnLt0?si=YRjrl2U0waLkNWqm&t=2151  ~MATH-X
 +
==Video Solution (Solve under 60 seconds!!!)==
 +
https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=588
 +
 
 +
~hsnacademy
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/rPRis7sGroI
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution (Animated)==
 
https://youtu.be/NivfOThj1No
 
https://youtu.be/NivfOThj1No
  
 
~Star League (https://starleague.us)
 
~Star League (https://starleague.us)
 +
 +
==Video Solution by Magic Square==
 +
https://youtu.be/-N46BeEKaCQ?t=4439
 +
==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=1299
 +
 +
==Video Solution by harungurcan==
 +
https://www.youtube.com/watch?v=VqN7c5U5o98&t=16s
 +
 +
~harungurcan
 +
 +
==See Also==
 +
{{AMC8 box|year=2023|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Revision as of 01:11, 21 August 2024

Problem

Along the route of a bicycle race, $7$ water stations are evenly spaced between the start and finish lines, as shown in the figure below. There are also $2$ repair stations evenly spaced between the start and finish lines. The $3$rd water station is located $2$ miles after the $1$st repair station. How long is the race in miles? [asy] // Credits given to Themathguyd‎ and Kante314 usepackage("mathptmx"); size(10cm); filldraw((11,4.5)--(171,4.5)--(171,17.5)--(11,17.5)--cycle,mediumgray*0.4 + lightgray*0.6); draw((11,11)--(171,11),linetype("2 2")+white+linewidth(1.2)); draw((0,0)--(11,0)--(11,22)--(0,22)--cycle); draw((171,0)--(182,0)--(182,22)--(171,22)--cycle); draw((31,4.5)--(31,0)); draw((51,4.5)--(51,0)); draw((151,4.5)--(151,0)); label(scale(.85)*rotate(45)*"Water 1", (23,-13.5)); label(scale(.85)*rotate(45)*"Water 2", (43,-13.5)); label(scale(.85)*rotate(45)*"Water 7", (143,-13.5)); filldraw(circle((103,-13.5),.2)); filldraw(circle((98,-13.5),.2)); filldraw(circle((93,-13.5),.2)); filldraw(circle((88,-13.5),.2)); filldraw(circle((83,-13.5),.2)); label(scale(.85)*rotate(90)*"Start", (5.5,11)); label(scale(.85)*rotate(270)*"Finish", (176.5,11)); [/asy] $\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 96$

Solution

Suppose that the race is $d$ miles long. The water stations are located at \[\frac{d}{8}, \frac{2d}{8}, \ldots, \frac{7d}{8}\] miles from the start, and the repair stations are located at \[\frac{d}{3}, \frac{2d}{3}\] miles from the start.

We are given that $\frac{3d}{8}=\frac{d}{3}+2,$ from which \begin{align*} \frac{9d}{24}&=\frac{8d}{24}+2 \\ \frac{d}{24}&=2 \\ d&=\boxed{\textbf{(D)}\ 48}. \end{align*} ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM

Video Solution by Math-X (Let's first Understand the question)

https://youtu.be/Ku_c1YHnLt0?si=YRjrl2U0waLkNWqm&t=2151 ~MATH-X

Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=588

~hsnacademy

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/rPRis7sGroI

~Education, the Study of Everything

Video Solution (Animated)

https://youtu.be/NivfOThj1No

~Star League (https://starleague.us)

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=4439

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=1299

Video Solution by harungurcan

https://www.youtube.com/watch?v=VqN7c5U5o98&t=16s

~harungurcan

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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