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Difference between revisions of "2023 AMC 12A Problems/Problem 1"

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==Problem 1==
 
==Problem 1==
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Cities <math>A</math> and <math>B</math> are <math>45</math> mile apart. Alicia lives in <math>A</math> and Beth lives in <math>B</math>. Alicia bikes towards <math>B</math> at 18 miles per hour. Leaving at the same time, Beth bikes toward <math>A</math> at 12 miles per hour. How many miles from City <math>A</math> will they be when they meet?
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Cities <math>A</math> and <math>B</math> are <math>45</math> miles apart. Alicia lives in <math>A</math> and Beth lives in <math>B</math>. Alicia bikes towards <math>B</math> at 18 miles per hour. Leaving at the same time, Beth bikes toward <math>A</math> at 12 miles per hour. How many miles from City <math>A</math> will they be when they meet?
  
 
<math>\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math>
 
<math>\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math>

Revision as of 14:45, 13 September 2024

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 $18\times1.5=\boxed{\textbf{(E) 27}}$

~zhenghua

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 $x=18\cdot\frac{3}{2}=\boxed{\textbf{(E) 27}}$

~walmartbrian

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. $\boxed{\textbf{(E) 27}}$

~Dilip

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.

~uwusugardaddy14

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)

Solution 8

2023 AMC12B Q1 P1.png
2023AMC12B Q1 P2.png

Note that Alicia and Beth must have travelled 45 miles together in order for them to meet together. While travelling the 45 miles, Alicia came closer towards Beth and Beth came closer towards Alicia. Since Alicia is travelling faster than Beth, we know that they will meet slightly closer towards city B from the middle. Also, the distance remaining between Alicia and Beth as they bike toward each other is proportional to their combined velocity. The combined velocity is $18t + 12t = 30t$. The time it takes to travel 45 miles going at 30 miles per hour can be calculated using simple algebra.

Let $t$ be the time it takes for Alicia and Beth to meet together.

\begin{align*} 18t + 12t = 45 \\ 30t = 45 \\ t = \frac{3}{2} \end{align*}

It takes $\frac{3}{2}$ hours of cycling by both Alicia and Beth at their respective biking speeds in order to meet together. Also, note that the question asks for the distance from City A they will be when they meet. This is the same as the distance travelled by Alicia from City A. Alicia travels at 18 mph, which turns into a distance = speed x time question.

\begin{align*} d &= st \\ d &= 18 \times \frac{3}{2} \\ d &= 27 \end{align*} The distance from City A when Alicia and Beth meet is $\boxed{\textbf{(E)}\ 27}$.

Note - Even though it might seem like there is a lot of work for this problem, it would typically take a very short time to think this entire process out.

~S. Khunti

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

https://youtu.be/GP-DYudh5qU?si=TdTUYY4j-g_K_8-b&t=115 ~Math-X


Video Solution by Power Solve (clear)

https://youtu.be/YXIH3UbLqK8?si=YFt4h7wm7XHun1lK

Video Solution 1 (⚡Under 1 min⚡)

https://youtu.be/7p1veMfoZO4

~Education, the Study of Everything


Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=WkRt_n1TEKY

Video Solution

https://youtu.be/eOialvQRL9Q

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

Video Solution (easy to digest) by Power Solve

https://www.youtube.com/watch?v=8huvzWTtgaU

Video Solution by DR.GOOGLE (YT: Pablo's Math)

https://youtu.be/AI6QlgxGVls

See also

2023 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
2023 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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