2023 AMC 8 Problems/Problem 13

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.

Animated Video Solution

https://youtu.be/NivfOThj1No

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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.

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2023 AMC 8 Problems/Problem 13

Animated Video Solution

Written Solution