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Difference between revisions of "2023 AMC 8 Problems/Problem 13"
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==Problem==
  
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Along the route of a bicycle race, 7 water stations are evenly spaced between the start and finish lines,
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as shown in the figure below. There are also 2 repair stations evenly spaced between the start and
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finish lines. The 3rd water station is located 2 miles after the 1st repair station. How long is the race
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in miles?
  
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==Written Solution==
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[[File:2023 AMC 8-13.png|thumb|center|300px]]
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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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==Solution==
  
 
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.
 
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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==Animated Video Solution==
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==Video Solution (Animated)==
 
https://youtu.be/NivfOThj1No
 
https://youtu.be/NivfOThj1No
  
 
~Star League (https://starleague.us)
 
~Star League (https://starleague.us)

Revision as of 05:52, 25 January 2023

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 3rd water station is located 2 miles after the 1st repair station. How long is the race in miles?

2023 AMC 8-13.png

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 96$

Solution

Knowing that there are $7$ equally spaced water stations they are each located $\frac{d}{8}$, $\frac{2d}{8}$,… $\frac{7d}{8}$ of the way from the start. Using the same logic for the $3$ station we have $\frac{d}{3}$ and $\frac{2d}{3}$ for the repair stations. It is given that the 3rd water is $2$ miles ahead of the $1$st repair station. So setting an equation we have $\frac{3d}{8} = \frac{d}{3} + 2$ getting common denominators $\frac{9d}{24} = \frac{8d}{24} + 2$ so then we have $d = \boxed{\text{(D)}48}$ from this.

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat


Video Solution (Animated)

https://youtu.be/NivfOThj1No

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

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