Difference between revisions of "2023 AMC 12A Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
− | This is a | + | This is a <math>d=st</math> problem, so let <math>x</math> be the time it takes to meet. We can write the following equation: |
<cmath>12x+18x=45</cmath> | <cmath>12x+18x=45</cmath> | ||
Solving gives us <math>x=1.5</math>. The <math>18x</math> is Alicia so <math>18\times1.5=\boxed{\textbf{(E) 27}}</math> | Solving gives us <math>x=1.5</math>. The <math>18x</math> is Alicia so <math>18\times1.5=\boxed{\textbf{(E) 27}}</math> | ||
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==Solution 2== | ==Solution 2== | ||
− | + | The relative speed of the two is <math>18+12=30</math>, so <math>\frac{3}{2}</math> hours would be required to travel <math>45</math> miles. <math>d=st</math>, so <math>x=18\cdot\frac{3}{2}=\boxed{\textbf{(E) 27}}</math> | |
~walmartbrian | ~walmartbrian |
Revision as of 22:04, 9 November 2023
Contents
[hide]Problem
Cities and are miles apart. Alicia lives in and Beth lives in . Alicia bikes towards at 18 miles per hour. Leaving at the same time, Beth bikes toward at 12 miles per hour. How many miles from City will they be when they meet?
Solution 1
This is a problem, so let be the time it takes to meet. We can write the following equation: Solving gives us . The is Alicia so
~zhenghua
Solution 2
The relative speed of the two is , so hours would be required to travel miles. , so
~walmartbrian
Solution 3
Since mph is times mph, Alicia will travel times as far as Beth. If is the distance Beth travels, Since this is the amount Beth traveled, the amount that Alicia traveled was
~daniel luo
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.