Difference between revisions of "2023 AMC 12A Problems/Problem 1"
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==Problem== | ==Problem== | ||
Cities <math>A</math> and <math>B</math> are <math>45</math> miles apart. Alicia lives in <math>A</math> and Beth lives in <math>B</math>. Alicia bikes towards <math>B</math> at 18 miles per hour. Leaving at the same time, Beth bikes toward <math>A</math> at 12 miles per hour. How many miles from City <math>A</math> will they be when they meet? | Cities <math>A</math> and <math>B</math> are <math>45</math> miles apart. Alicia lives in <math>A</math> and Beth lives in <math>B</math>. Alicia bikes towards <math>B</math> at 18 miles per hour. Leaving at the same time, Beth bikes toward <math>A</math> at 12 miles per hour. How many miles from City <math>A</math> will they be when they meet? |
Revision as of 10:53, 10 November 2023
- The following problem is from both the 2022 AMC 10A #1 and 2022 AMC 12A #1, so both problems redirect to this page.
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
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.