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

Revision as of 20:25, 24 January 2023

Animated Video Solution

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

Revision as of 20:06, 24 January 2023 (view source) Sohumuttamchandani (talk \| contribs) (→‎Animated Video Solution) ← Older edit		Revision as of 20:25, 24 January 2023 (view source) Apex304 (talk \| contribs) Newer edit →
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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.

	==Animated Video Solution==		==Animated Video Solution==

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

Revision as of 20:25, 24 January 2023

Animated Video Solution

Written Solution