Difference between revisions of "2022 AMC 10A Problems/Problem 2"
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<math>\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13</math> | <math>\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13</math> | ||
| − | == Solution == | + | == Solution 1== |
Mike's speed is <math>\frac{15}{57}=\frac{5}{19}</math> laps per minute. | Mike's speed is <math>\frac{15}{57}=\frac{5}{19}</math> laps per minute. | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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| + | == Solution 2== | ||
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| + | Mike runs <math>1</math> lap in <math>\frac{57}{15}=\frac{19}{5}</math> minutes. So, in <math>27</math> minutes, Mike ran about <math>\frac{27}{\frac{19}{5}} \approx \boxed{\textbf{(B) }7}</math> laps. | ||
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| + | ~MrThinker | ||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== | ||
Revision as of 15:39, 14 November 2022
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
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. 
