Art of Problem Solving

2025 AMC 10A Problems/Problem 1: Difference between revisions

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== Problem ==
#redirect [[2025 AMC 12A Problems/Problem 1]]
 
Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at <imath>1:30</imath>, traveling due northat a steady <imath>8</imath> mile per hour. Betsy leaves on her bicycle from the same point at <imath>2:30</imath>, traveling due east at a steady <imath>12</imath> miles per hour. At what time will thy be exactly the same distance from their common starting point?
 
<imath>\textbf{(A) } {3:30}\qquad\textbf{(B) } {3:45}\qquad\textbf{(C) } {4:00}\qquad\textbf{(D) } {4:15}\qquad\textbf{(E) } {4:30}</imath>
 
 
== Solution==
We can see that Betsy travles 1 hour after Andy started. We have <imath>lcm(8, 12)=24</imath> now we can find the time traveled \(\frac{24}{8} = 3 \text{ hours}\)
 
Now we have time \(1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}\)
 
-Boywithnuke(Goal: 10 followers)

Latest revision as of 02:13, 8 November 2025