2023 AMC 12A Problems/Problem 1
- The following problem is from both the 2023 AMC 10A #1 and 2023 AMC 12A #1, so both problems redirect to this page.
Contents
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
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.
~Dilip
==Solution 5 (Under 20 seconds) We know that Alice approaches Beth at mph and Beth approaches Alice at mph. If we consider that if Alice moves miles at the same time Beth moves miles and then more miles at the same time Alice moves more miles then Alice has moved miles from point A at the same time that Beth has moved miles from point B meaning that Alice and Beth meet 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
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.