Difference between revisions of "2023 AMC 12A Problems/Problem 1"
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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== | ||
+ | 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.
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 —> miles left. Alice then moves more miles at the same time as Beth moves more miles. 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 ~SP
Solution 6 (simple linear equations)
We know that Beth starts 45 miles away from City A, let’s create two equations: Alice-> Beth-> [-12 is the slope; 45 is the y-intercept]
Solve the system:
So,
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 , which is equivalent to the ratio of their distances. Therefore, Alicia travels of their combined speed and thus their combined distance, meaning that she travels miles.
-Benedict T (countmath1)
Video Solution 1 (⚡Under 1 min⚡)
~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
~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 (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.