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

(Video Solution by Math-X (First understand the problem!!!))
(Solution 6 (simple linear equations))
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So, <math>18(1.5)=</math> <math>\boxed{\textbf{(E) 27}}</math>
 
So, <math>18(1.5)=</math> <math>\boxed{\textbf{(E) 27}}</math>
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=Solution 7==
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Since Alicia and Beth's speeds are constant, they are proportional to their distances covered. The ratio of Alicia's speed to Beth's speed is <math>18:12 = 3:2</math>, which is equivalent to the ratio of their distances. Therefore, Alicia travels <math>\frac{3}{5}</math> of their combined speed and thus their combined distance, meaning that she travels <math>\frac{3}{5}\cdot 45 = \boxed{\textbf{(E)}\ 27}</math> miles.
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-Benedict T (countmath1)
  
 
==Video Solution 1 (⚡Under 1 min⚡)==
 
==Video Solution 1 (⚡Under 1 min⚡)==

Revision as of 14:09, 29 January 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

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.

~MC_ADe ~SP

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 proportional to their distances covered. The ratio of Alicia's speed to Beth's speed is $18:12 = 3:2$, which is equivalent to the ratio of their distances. Therefore, Alicia travels $\frac{3}{5}$ of their combined speed and thus their combined distance, meaning that she travels $\frac{3}{5}\cdot 45 = \boxed{\textbf{(E)}\ 27}$ miles.


-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

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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