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

Latest revision as of 20:24, 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

@@ Line 6: / Line 6: @@
 <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>
+==This page have some problum with rendering, thx for everyone who is trying to fix it==
 == Solution 1 ==
@@ Line 37: / Line 41: @@
 At E, 4:30, we see that both Andy and Betsy have gone 24 miles.
-Now we see that <imath>\textbf{(E) } {4{:}30}</imath> is the correct answer.
+Now we see that \boxed{\textbf{(E) } 4{:}30}<imath> is the correct answer.
 ~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>
+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>
+So the answer is </imath>1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}<imath>
 OR
@@ Line 57: / Line 61: @@
 ==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>.
+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)
+~minor </imath>\LaTeX<imath> edits by i_am_not_suk_at_math (saharshdevaraju 21:17, 6 November 2025 (EST)saharshdevaraju)
 ==Solution 6 - Very Simple==
@@ Line 76: / Line 80: @@
 <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>.
+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)
@@ Line 88: / Line 92: @@
 OVERKILL using Calculus:
-We define a function <imath>F(t)</imath> that represents the <imath>\textbf{difference}</imath> in their distances:
+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:
@@ Line 94: / Line 98: @@
 <cmath>F(t) = 2t - 6</cmath>
-The time when their distances are equal is the time when <imath>F(t) = 0</imath>.
+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>:
+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>.
+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:
@@ Line 110: / Line 114: @@
-This calculus verification confirms that the unique solution occurs at <imath>t=3</imath> hours.
+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}</imath>:
+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>
+<cmath>\text{Time} = 1{:}30 \text{ PM} + 3 \text{ hours} = 4{:}30 \text{ PM}</cmath>
 -apex304
 On
 == Video Solution (Fast and Easy) ==
 https://youtu.be/fjpeOuUMOyc?si=ROuzlwgz7UByUx_9

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