Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Video Solution (easy to digest) by Power Solve)
(29 intermediate revisions by 16 users not shown)
Line 1: Line 1:
−
==Problem==
+
{{duplicate|[[2023 AMC 10A Problems/Problem 1|2023 AMC 10A #1]] and [[2023 AMC 12A Problems/Problem 1|2023 AMC 12A #1]]}}
 +
 
 +
==Problem 1==
 
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?
 
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>
Line 24: Line 26:
  
 
~Dilip
 
~Dilip
 +
 +
==Solution 5 (Under 20 seconds)==
 +
We know that Alice approaches Beth at <math>18</math> mph and Beth approaches Alice at <math>12</math> mph. If we consider that if Alice moves <math>18</math> miles at the same time Beth moves <math>12</math> miles —> <math>15</math> miles left. Alice then moves <math>9</math> more miles at the same time as Beth moves <math>6</math> more miles. Alice has moved <math>27</math> miles from point A at the same time that Beth has moved <math>18</math> miles from point B, meaning that Alice and Beth meet <math>27</math> 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-> <math>18t=d</math>
 +
Beth->  <math>-12t+45=d</math> [-12 is the slope; 45 is the y-intercept]
 +
 +
Solve the system:
 +
 +
<math>18t=-12t+45
 +
30t=45
 +
t=1.5</math>
 +
 +
So, <math>18(1.5)=</math> <math>\boxed{\textbf{(E) 27}}</math>
 +
 +
=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 <math>\frac{18}{30} = \frac{3}{5}</math> of their combined speed, she travels <math>\frac{3}{5}\cdot 45 = \boxed{\textbf{(E)}\ 27}</math> of the total distance.
 +
 +
-Benedict T (countmath1)
 +
 +
==Video Solution 1 (⚡Under 1 min⚡)==
 +
https://youtu.be/7p1veMfoZO4
 +
 +
~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🔥]==
 +
 +
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==
 
==See also==
 +
{{AMC10 box|year=2023|ab=A|before=First Problem|num-a=2}}
 
{{AMC12 box|year=2023|ab=A|before=First Problem|num-a=2}}
 
{{AMC12 box|year=2023|ab=A|before=First Problem|num-a=2}}
−
{{AMC10 box|year=2023|ab=A|before=First Problem|num-a=2}}
 
  
 
[[Category:Rate Problems]]
 
[[Category:Rate Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:57, 5 May 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)

Video Solution 1 (⚡Under 1 min⚡)

https://youtu.be/7p1veMfoZO4

~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🔥]

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_1&oldid=219039"