Difference between revisions of "2023 AMC 8 Problems/Problem 13"
(Ayo, get back to grinding stop trying to cheat. Become one with the Sigma :).) |
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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. | |
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Revision as of 18:28, 24 January 2023
Knowing that there are 
equally spaced water stations they are each located 
, 
,… 
of the way from the start. Using the same logic for the 
station we have 
and 
for the repair stations. It is given that the 3rd water is 
miles ahead of the 
st repair station. So setting an equation we have 
getting common denominators 
so then we have 
from this.
