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2025 AMC 10A Problem 1

Problem 1 of 25EasierAlgebraArithmetic

Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at 1:30,1{:}30, traveling due north at a steady 88 miles per hour. Betsy leaves on her bicycle from the same point at 2:30,2{:}30, traveling due east at a steady 1212 miles per hour. At what time will they be exactly the same distance from their common starting point?

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Solution

Let tt be the hours since 1:30.1{:}30. Andy has gone 8t8t miles north. Betsy starts an hour later, so she’s gone 12(t−1)12(t-1) miles east. We want these equal: 8t=12(t−1).8t = 12(t-1). That gives 4t=12,4t = 12, so t=3.t = 3. Three hours past 1:301{:}30 is 4:30.4{:}30. Thus, E is the correct answer.
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Tagged: distance rate and time · linear equation

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