2023 AMC 12A Problems/Problem 1

The following problem is from both the 2023 AMC 10A #1 and 2023 AMC 12A #1, so both problems redirect to this page.

Problem 1

Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$. Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$

Solution 1

This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$. The $18x$ is Alicia so $18\times1.5=\boxed{\textbf{(E) 27}}$


Solution 2

The relative speed of the two is $18+12=30$, so $\frac{3}{2}$ hours would be required to travel $45$ miles. $d=st$, so $x=18\cdot\frac{3}{2}=\boxed{\textbf{(E) 27}}$


Solution 3

Since $18$ mph is $\frac{3}{2}$ times $12$ mph, Alicia will travel $\frac{3}{2}$ times as far as Beth. If $x$ is the distance Beth travels, \[\frac{3}{2}x+x=45\] \[\frac{5}{2}x=45\] \[x=18\]Since this is the amount Beth traveled, the amount that Alicia traveled was \[45-18=\boxed{\textbf{(E) 27}}\]

~daniel luo

Solution 4

Alice and Barbara close in on each other at 30mph. Since they are 45 miles apart, they will meet in t = d/s = 45miles / 30mph = 3/2 hours. We can either calculate the distance Alice travels at 18mph or the distance Barbara travels at 12mph; since we want the distance from Alice, we go with the former. Alice (and Barbara) will meet in 1 1/2 hours at 18mph x 3/2 hours = 27 miles from A. $\boxed{\textbf{(E) 27}}$


Solution 5 (Under 20 seconds)

We know that Alice approaches Beth at $18$ mph and Beth approaches Alice at $12$ mph. If we consider that if Alice moves $18$ miles at the same time Beth moves $12$ miles —> $15$ miles left. Alice then moves $9$ more miles at the same time as Beth moves $6$ more miles. Alice has moved $27$ miles from point A at the same time that Beth has moved $18$ miles from point B, meaning that Alice and Beth meet $27$ miles from point A.


Solution 6 (simple linear equations)

We know that Beth starts 45 miles away from City A, let’s create two equations: Alice-> $18t=d$ Beth-> $-12t+45=d$ [-12 is the slope; 45 is the y-intercept]

Solve the system:

$18t=-12t+45 30t=45 t=1.5$

So, $18(1.5)=$ $\boxed{\textbf{(E) 27}}$

Solution 7

Since Alicia and Beth's speeds are constant, they are directly proportional to their distances covered, so the ratio of their speeds is equal to the ratio of their covered distances. Since Alicia travels $\frac{18}{30} = \frac{3}{5}$ of their combined speed, she travels $\frac{3}{5}\cdot 45 = \boxed{\textbf{(E)}\ 27}$ of the total distance.

-Benedict T (countmath1)

Video Solution 1 (⚡Under 1 min⚡)


~Education, the Study of Everything

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=Ngh2w5-AAuP38GZk&t=34 ~Math-X

Video Solution by CosineMethod [🔥Fast and Easy🔥]


Video Solution


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution (easy to digest) by Power Solve


See also

2023 AMC 10A (ProblemsAnswer KeyResources)
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
2023 AMC 12A (ProblemsAnswer KeyResources)
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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