2025 AMC 10A Problems/Problem 1

The following problem is from both the 2025 AMC 10A #1 and 2025 AMC 12A #1, so both problems redirect to this page.

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 1

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 \(1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}\)

-Boywithnuke(Goal: 10 followers)

Solution 2

hours after Betsy left, Andy has traveled miles, and Betsy has traveled miles. We are told these are equal, so . Solving, we get , so Andy and Betsy will be exactly the same distance from their common starting point two hours after Betsy leaves, or .

~mithu542

Solution 3

We can just use all the answer choices that we are given.

Let's use casework for each of the answers:

At 3:30, Andy will have gone 2*8=16 miles. Betsy will have gone 1*12=12 miles. At 3:45, we can just add the distance each of them goes in 1/4 hours. Therefore, Andy goes 18 miles, and Betsy goes 15 miles. At 4:00, we see from the same logic that Andy has gone 20 miles, and Betsy has gone 18 miles. At 4:15 we see that Andy has gone 22 miles, and Betsy has gone 21 miles. At E, 4:30, we see that both Andy and Betsy have gone 24 miles.

No we see that . is the correct answer.

~vgarg

Solution 4

We can see that at , Andy will be miles ahead. For every hour that they both travel, Betsy will gain miles on . Therefore, it will take more hours for Betsy to catch up, and they will be at the same point at .