2025 AMC 10A Problems/Problem 1: Difference between revisions
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== Problem == | == Problem == | ||
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 | 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 they 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> | <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== | == Solution== | ||
Revision as of 12:44, 6 November 2025
Problem
Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at 
, traveling due northat a steady 
mile per hour. Betsy leaves on her bicycle from the same point at 
, traveling due east at a steady 
miles per hour. At what time will they be exactly the same distance from their common starting point?
Solution
We can see that Betsy travles 1 hour after Andy started. We have 
now we can find the time traveled \(\frac{24}{8} = 3 \text{ hours}\)
Now we have time \(1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}\)
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