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

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~walmartbrian
 
~walmartbrian
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==Solution 3==
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Since <math>18</math> mph is <math>\frac{3}{2}</math> times <math>12</math> mph, Alicia will travel <math>\frac{3}{2}</math> times as far as Beth. If <math>x</math> is the distance Beth travels, <cmath>\frac{3}{2}x+x=45</cmath> <cmath>\frac{5}{2}x=45</cmath> <cmath>x=18</cmath>Since this is the amount Beth traveled, the amount that Alicia traveled was <cmath>45-18=\boxed{\textbf{(E) 27}}</cmath>
  
 
==See also==
 
==See also==

Revision as of 20:46, 9 November 2023

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 Distance=Time$\times$Speed 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

We can solve this by first adding 18 and 12 mph, which gets us 30mph. Their collective speed is 30mph, which means they will mean when they travel 45 miles at 30mph which would take 90 minutes. Then 18 miles per hour for 90 minutes is 27 miles. (E)

~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 \[45-18=\boxed{\textbf{(E) 27}}\]

See also

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
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

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