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2025 AMC 12A Problems/Problem 1: Difference between revisions

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At E, 4:30, we see that both Andy and Betsy have gone 24 miles.  
At E, 4:30, we see that both Andy and Betsy have gone 24 miles.  


Now we see that \boxed{\textbf{(E) } 4{:}30}$ is the correct answer.  
Now we see that \boxed{\textbf{(E) } 4{:}30}<imath> is the correct answer.  


~vgarg
~vgarg
==Solution 4==
We can see that Betsy travels 1 hour after Andy started. We have </imath>lcm (8, 12)=24<imath>. Now we can find the total time Andy has taken once Betsy catches up: </imath>\frac{24}{8} = 3 \text{ hours}<imath>
So the answer is </imath>1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}<imath>
OR
You can simply write it out. After one hour, where will Andy be located and where will Betsy be located based on their times?
~Boywithnuke(Goal: 10 followers)
~minor edits by ChickensEatGrass
~minor edits by RISHADA
==Solution 5==
The distance Andy travels can be represented by </imath>8x<imath> and Betsy with the equation </imath>12(x-1).<imath> The solution to this is </imath>x = 3,<imath> so the answer is </imath>3<imath> hours after </imath>1:30<imath> therefore,the solution is </imath>\textbf{(E) } {4{:}30}<imath>.
~minor LaTeX edits by zoyashaikh
~minor </imath>\LaTeX<imath> 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.
<cmath></cmath>
\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}
<cmath></cmath>
Based on this very simple table, we can conclude that they will be the exact same difference from their mutual starting point at </imath>\textbf{(E) } {4{:}30}<imath>.
~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 </imath>F(t)<imath> that represents the </imath>\textbf{difference}<imath> in their distances:
<cmath>F(t) = D_B(t) - D_A(t)</cmath>
Substituting the distance functions:
<cmath>F(t) = (12t - 6) - (10t) \quad \text{for } t \ge 0.5</cmath>
<cmath>F(t) = 2t - 6</cmath>
The time when their distances are equal is the time when </imath>F(t) = 0<imath>.
We take the first derivative of </imath>F(t)<imath> with respect to time </imath>t<imath>:
<cmath>F'(t) = \frac{d}{dt} F(t) = \frac{d}{dt} (2t - 6)</cmath>
<cmath>F'(t) = 2</cmath>
Since </imath>F'(t) = 2 > 0<imath>, the difference function </imath>F(t)<imath> is strictly increasing. This confirms that the point where </imath>F(t) = 0<imath> is a \textbf{unique root}, meaning there is only one moment in time when </imath>D_A(t) = D_B(t)<imath>.
We set the difference function to zero:
<cmath>F(t) = 0</cmath>
<cmath>2t - 6 = 0</cmath>
<cmath>2t = 6</cmath>
<cmath>t = 3 \text{ hours}</cmath>
This calculus verification confirms that the unique solution occurs at </imath>t=3<imath> hours.
The time is </imath>3<imath> hours after </imath>1{:}30 \text{ PM}$:
<cmath>\text{Time} = 1{:}30 \text{ PM} + 3 \text{ hours} = 4{:}30 \text{ PM}</cmath>
-apex304
On


== Video Solution (Fast and Easy) ==
== Video Solution (Fast and Easy) ==

Revision as of 20:20, 11 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 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?

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


This page have some problum with rendering, thx for everyone who is trying to fix it

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 $\textbf{(E) } {4{:}30}$.


~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 \boxed{\textbf{(E) } 4{:}30}$is the correct answer.

~vgarg

==Solution 4== We can see that Betsy travels 1 hour after Andy started. We have$ (Error compiling LaTeX. Unknown error_msg)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$1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}$OR

You can simply write it out. After one hour, where will Andy be located and where will Betsy be located based on their times?

~Boywithnuke(Goal: 10 followers)

~minor edits by ChickensEatGrass

~minor edits by RISHADA

==Solution 5== The distance Andy travels can be represented by$ (Error compiling LaTeX. Unknown error_msg)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$\textbf{(E) } {4{:}30}$.

~minor LaTeX edits by zoyashaikh

~minor$ (Error compiling LaTeX. Unknown error_msg)\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. <cmath></cmath> \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} <cmath></cmath>

Based on this very simple table, we can conclude that they will be the exact same difference from their mutual starting point at$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)F(t)$that represents the$\textbf{difference}$in their distances: <cmath>F(t) = D_B(t) - D_A(t)</cmath> Substituting the distance functions: <cmath>F(t) = (12t - 6) - (10t) \quad \text{for } t \ge 0.5</cmath> <cmath>F(t) = 2t - 6</cmath>

The time when their distances are equal is the time when$ (Error compiling LaTeX. Unknown error_msg)F(t) = 0$.

We take the first derivative of$ (Error compiling LaTeX. Unknown error_msg)F(t)$with respect to time$t$: <cmath>F'(t) = \frac{d}{dt} F(t) = \frac{d}{dt} (2t - 6)</cmath> <cmath>F'(t) = 2</cmath>

Since$ (Error compiling LaTeX. Unknown error_msg)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: <cmath>F(t) = 0</cmath> <cmath>2t - 6 = 0</cmath> <cmath>2t = 6</cmath> <cmath>t = 3 \text{ hours}</cmath>


This calculus verification confirms that the unique solution occurs at$ (Error compiling LaTeX. Unknown error_msg)t=3$hours.

The time is$ (Error compiling LaTeX. Unknown error_msg)3$hours after$1{:}30 \text{ PM}$: \[\text{Time} = 1{:}30 \text{ PM} + 3 \text{ hours} = 4{:}30 \text{ PM}\]

-apex304

On

Video Solution (Fast and Easy)

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

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 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

See Also

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

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