2023 AMC 10A Problem 1
Cities and are miles apart. Alicia lives in and Beth lives in . Alicia bikes towards at miles per hour. Leaving at the same time, Beth bikes toward at miles per hour. How many miles from City will they be when they meet?
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Key insight
Each hour the two riders get 18 + 12 = 30 miles closer, so they meet after 1.5 hours. Then find how far Alicia rode.
Solution
1. Look at one hour first. Alicia rides miles toward Beth and Beth rides miles toward Alicia. So the gap between them shrinks by miles every hour.
2. The gap starts at miles. It takes
for the gap to close, so they meet after hours.
3. The question asks for the distance from City . Alicia starts at , so we need how far she rode: miles.
4. Check with Beth: she rides miles, and , the full distance.
The answer is .
Why this works
When two people move toward each other, you can pretend one of them stands still while the other moves at the two speeds added together. That gives the meeting time, and then distance = speed × time for each rider. The last step is to reread the question: it asks for the distance from , which is Alicia's distance, not Beth's.
Alternative approach
Both ride for the same amount of time, so the distances they cover are in the same ratio as their speeds, . Split miles into equal parts of miles. Alicia gets parts, which is miles.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Since the two riders move toward each other, the distance between them shrinks at miles per hour. The -mile gap closes in
In that time Alicia, who starts at , rides miles. So they meet miles from .
Check: Beth rides miles, and .
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
- Introduction to Algebra, Chapter 6: Ratios and Percents
- Introduction to Algebra, Chapter 7: Proportion (§7.4)
- Prealgebra, Chapter 13: Data and Statistics (§13.1)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.