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 $1{:}30$ , traveling due northat a steady $8$ mile per hour. Betsy leaves on her bicycle from the same point at $2{:}30$ , traveling due east at a steady $12$ miles per hour. At what time will they be exactly the same distance from their common starting point?

$\textbf{(A) } {3:30}\qquad\textbf{(B) } {3:45}\qquad\textbf{(C) } {4:00}\qquad\textbf{(D) } {4:15}\qquad\textbf{(E) } {4:30}$

Solution 1

We can see that Betsy travels 1 hour after Andy started. We have $lcm(8, 12)=24$ . Now we can find the total time Andy has taken once Betsy catches up: $\frac{24}{8} = 3 \text{ hours}$

So the answer is $1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}$

~Boywithnuke(Goal: 10 followers)

~minor edits by ChickensEatGrass

Solution 2

a. $h$ hours after Betsy left, Andy has traveled $8(h+1)$ miles, and Betsy has traveled $12h$ miles. We are told these are equal, so $8h+8=12h$ . Solving, we get $h=2$ , so Andy and Betsy will be exactly the same distance from their common starting point two hours after Betsy leaves, or $\text{(E) }4:30$ .

b. Let $h$ be the number of hours after Betsy starts. Then Andy has been traveling for $h+1$ hours, so he has gone $8(h+1)$ miles, and Betsy has traveled $12h$ miles. Setting these equal, we get $8(h+1) = 12h$ , which simplifies to $8h + 8 = 12h$ , so $4h = 8$ and $h = 2$ . Thus, Betsy catches up 2 hours after she starts. Since Andy started 1 hour earlier at 1:30, the total time from Andy’s start is $h+1 = 3$ hours, giving a catch-up time of $1:30 + 3 = 4:30$ . Answer: $\text{(E) }4:30$ .

~mithu542 ( 2a) ~kapiltheangel ( 2b)

Solution 3 (bash)

We can 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 $2\cdot8=16$ miles. Betsy will have gone $1\cdot12=12$ 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.

Now we see that $\text{(E) }4:30$ is the correct answer.

~vgarg

Solution 4

We can see that at $2:30$ , Andy will be $8$ miles ahead. For every hour that they both travel, Betsy will gain $4$ miles on Andy. Therefore, it will take $2$ more hours for Betsy to catch up, and they will be at the same point at $\text{(E) }4:30$ .

~vinceS

Solution 5

The distance Andy travels can be represented by $8x$ and Betsy with the equation $12(x-1).$ The solution to this is $x = 3,$ so the answer is $1:30$ plus $3$ hours or $\boxed{\text{(E) }4:30}$ .

~minor LaTeX edits by zoyashaikh

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search