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 Andy. Therefore, it will take more hours for Betsy to catch up, and they will be at the same point at .

~vinceS