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| == Problem ==
| | #redirect [[2025 AMC 12A Problems/Problem 1]] |
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| 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?
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| <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>
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| == Solution 1 ==
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| 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 \(\frac{24}{8} = 3 \text{ hours}\)
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| Now we have time \(1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}\)
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| -Boywithnuke(Goal: 10 followers)
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| == Solution 2 ==
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| <imath>h</imath> hours after Betsy left, Andy has traveled <imath>8(h+1)</imath> miles, and Betsy has traveled <imath>12h</imath> miles. We are told these are equal, so <imath>8h+8=12h</imath>. Solving, we get <imath>h=2</imath>, so Andy and Betsy will be exactly the same distance from their common starting point two hours after Betsy leaves, or <imath>\text{(E) }4:30</imath>.
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| ~mithu542
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