2025 AMC 10A Problems/Problem 1: Difference between revisions

Revision as of 21:54, 6 November 2025

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 that 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

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 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 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

Using the information that they move at a constant rate, we can create a small table based on their place (in miles) by hours. $\[\]$ \begin{array}{c|c|c} \text{-} & \text{Andy} & \text{Betsy} \\[6pt] \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

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 Daily Dose of Math

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

@@ Line 3: / Line 3: @@
 == Problem ==
-Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at <imath>1{:}30</imath>, traveling due northat a steady <imath>8</imath> mile per hour. Betsy leaves on her bicycle from the same point at <imath>2{:}30</imath>, traveling due east at a steady <imath>12</imath> miles per hour. At what time will they be exactly the same distance from their common starting point?
+Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at <imath>1{:}30</imath>, traveling due north that a steady <imath>8</imath> mile per hour. Betsy leaves on her bicycle from the same point at <imath>2{:}30</imath>, traveling due east at a steady <imath>12</imath> miles per hour. At what time will they be exactly the same distance from their common starting point?
 <imath>\textbf{(A) } {3:30}\qquad\textbf{(B) } {3:45}\qquad\textbf{(C) } {4:00}\qquad\textbf{(D) } {4:15}\qquad\textbf{(E) } {4:30}</imath>

2025 AMC 10A (Problems • Answer Key • Resources)
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