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Difference between revisions of "2023 AMC 8 Problems/Problem 13"

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==Solution==
 
==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. If this looks confusing, then think about it this way. If you examine the provided image, you will see that the road can be divided into eight equal sections. The first water station is <math>\frac{1}{8}</math> of the total distance, <math>d</math>. That gives us the fraction, <math>\frac{d}{8}</math>. This is how it leads to the rest of the fractions. 
+
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.
  
 
We are given that <math>\frac{3d}{8}=\frac{d}{3}+2,</math> from which   
 
We are given that <math>\frac{3d}{8}=\frac{d}{3}+2,</math> from which   
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d&=\boxed{\textbf{(D)}\ 48}.
 
d&=\boxed{\textbf{(D)}\ 48}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM, Srinjini_060313
+
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
  
 
==Video Solution by CoolMathProblems==
 
==Video Solution by CoolMathProblems==

Latest revision as of 18:39, 22 July 2025

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 CoolMathProblems

https://youtu.be/9WP3LQaMIVg?feature=shared&t=273

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

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

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=921 ~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

Video Solution by Dr. David

https://youtu.be/A7NZlithQ44

Video Solution by WhyMath

https://youtu.be/IdHONVZeyGo

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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