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

Revision as of 21:38, 10 November 2023

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

Problem

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 and then $9$ more miles at the same time Alice moves $6$ more miles then 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.

~MC_ADe

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

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

Video Solution

https://youtu.be/eOialvQRL9Q

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

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.

@@ Line 27: / Line 27: @@
 ~Dilip
-==Solution 5 (Under 20 seconds)
+==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 and then <math>9</math> more miles at the same time Alice moves <math>6</math> more miles then 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.

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