Difference between revisions of "2022 AMC 10A Problems/Problem 2"
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m (added solution 5 (approximation): This work is my own, and should not resemble other solutions found on the internet.) |
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~UltimateDL | ~UltimateDL | ||
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+ | == Solution 5 (approximation)== | ||
+ | <math>57</math> minutes is almost equal to <math>1</math> hour. Running <math>15</math> laps in <math>1</math> hour is running approximately <math>1</math> lap every <math>4</math> minutes. This means that in <math>27</math> minutes, Mike will run approximately <math>\frac{27}{4}</math> laps. This is very close to <math>\frac{28}{4} = 7</math>, so the answer is <math>\boxed{\textbf{(B) }7}</math>. | ||
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+ | ~TheGoldenRetriever | ||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== |
Revision as of 10:46, 15 July 2023
Contents
Problem
Mike cycled laps in minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first minutes?
Solution 1
Mike's speed is laps per minute.
In the first minutes, he completed approximately laps.
~MRENTHUSIASM
Solution 2
Mike runs lap in minutes. So, in minutes, Mike ran about laps.
~MrThinker
Solution 3
Mike's rate is where is the number of laps he can complete in minutes. If you cross multiply, .
So, .
~Shiloh000
Solution 4 (Quick Estimate)
Note that minutes is a little bit less than half of minutes. Mike will therefore run a little bit less than laps, which is about .
~UltimateDL
Solution 5 (approximation)
minutes is almost equal to hour. Running laps in hour is running approximately lap every minutes. This means that in minutes, Mike will run approximately laps. This is very close to , so the answer is .
~TheGoldenRetriever
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.