2025 AMC 12A 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 north at an 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 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 $\boxed{\textbf{(E) } 4{:}30}$ .

~vinceS

~minor LaTeX edits by zoyashaikh

Solution 2

You can look at this problem from both Andy's PoV and Betsy's PoV

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

Betsy(B). $h$ 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 $\text{(E) }4:30$ .

~kapiltheangel (2A) ~mithu542 (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 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 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 $3$ hours after $1:30$ therefore,the solution is $\boxed{\textbf{(E) }4:30}$ .

~minor LaTeX edits by zoyashaikh

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 21:17, 6 November 2025 (EST)saharshdevaraju)

Solution 6 - Very Simple

Using the information that they move at a constant rate, we can create a small table based on their place (in miles) by hours. Andy moves at 8 miles an hour, as seen below, and Betsy moves at 12 miles per hour after one hour after Andy begins, as shown in the table. $\[\]$ \begin{array}{c|c|c} \text{Time/Subject} & \text{Andy} & \text{Betsy} \\[6pt] \hline \text{1:30} & 0 & 0 \\[6pt] \text{2:30} & 8 & 0 \\[6pt] \text{3:30} & 16 & 12 \\[6pt] \text{4:30} & 24 & 24 \end{array} $\[\]$

Based on this very simple table, we can conclude that they will be the exact same difference from their mutual starting point at $\boxed{\textbf{(E) }4:30}$ .

~i_am_not_suk_at_math (saharshdevaraju 21:17, 6 November 2025 (EST)saharshdevaraju)

~minor LaTeX edits by vinceS

~minor minor edits by A-V-N-I

Solution 7

OVERKILL using Calculus:

We define a function $F(t)$ that represents the $\textbf{difference}$ in their distances: $F(t) = D_B(t) - D_A(t)$ Substituting the distance functions: $F(t) = (12t - 6) - (10t) \quad \text{for } t \ge 0.5$ $F(t) = 2t - 6$

The time when their distances are equal is the time when $F(t) = 0$ .

We take the first derivative of $F(t)$ with respect to time $t$ : $F'(t) = \frac{d}{dt} F(t) = \frac{d}{dt} (2t - 6)$ $F'(t) = 2$

Since $F'(t) = 2 > 0$ , the difference function $F(t)$ is strictly increasing. This confirms that the point where $F(t) = 0$ is a \textbf{unique root}, meaning there is only one moment in time when $D_A(t) = D_B(t)$ .

We set the difference function to zero: $F(t) = 0$ $2t - 6 = 0$ $2t = 6$ $t = 3 \text{ hours}$

This calculus verification confirms that the unique solution occurs at $t=3$ hours.

The time is $3$ hours after $1:30 \text{ PM}$ : $\text{Time} = 1:30 \text{ PM} + 3 \text{ hours} = 4:30 \text{ PM}$

-apex304

On

Video Solution by Power Solve

https://www.youtube.com/watch?v=QBn439idcPo

Chinese Video Solution

https://www.bilibili.com/video/BV1852uBoE8K/

~metrixgo

Video Solution (Intuitive, Quick Explanation!)

https://youtu.be/c-UDo53KwFU

~ Education, the Study of Everything

Video Solution (Fast and Easy)

https://youtu.be/fjpeOuUMOyc?si=ROuzlwgz7UByUx_9 ~ Pi Academy

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution by Daily Dose of Math 🔥🔥🔥

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video solution

https://youtu.be/l1RY_C20Q2M

Easy Solution

https://www.youtube.com/watch?v=kHwBHvvvTbY

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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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

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