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

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~Dilip
 
~Dilip
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/cMgngeSmFCY?si=Ngh2w5-AAuP38GZk&t=34 ~Math-X
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==See also==
 
==See also==

Revision as of 13:49, 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

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

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


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

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