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Difference between revisions of "2023 AMC 8 Problems/Problem 13"
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==Animated Video Solution==
 
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https://youtu.be/NivfOThj1No
 
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~Star League (https://starleague.us)
 
  
 
==Written Solution==
 
==Written Solution==
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
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==Animated Video Solution==
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https://youtu.be/NivfOThj1No
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~Star League (https://starleague.us)

Revision as of 19:37, 24 January 2023


Written 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


Animated Video Solution

https://youtu.be/NivfOThj1No

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

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