Difference between revisions of "2022 AMC 10A Problems/Problem 2"
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== Solution 4 (Reasonable Guess) == | == Solution 4 (Reasonable Guess) == | ||
| − | Note that 27 minutes is a little bit less than half of 57 minutes. Mike will therefore run a little bit less than <math>15/2=7.5</math> laps, which is about \boxed{\textbf{(B) }7} | + | Note that 27 minutes is a little bit less than half of 57 minutes. Mike will therefore run a little bit less than <math>15/2=7.5</math> laps, which is about <math>\boxed{\textbf{(B) }7}</math>. |
~UltimateDL | ~UltimateDL | ||
Revision as of 18:38, 21 November 2022
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 ![\[\frac{15}{57}=\frac{x}{27},\]](http://latex.artofproblemsolving.com/0/2/3/02357c82cc38c7308d0a04ce97912d5a1782a45a.png)
where 
is the number of laps he can complete in minutes.
If you cross multiply, 
.
So, 
.
~Shiloh000
Solution 4 (Reasonable Guess)
Note that 27 minutes is a little bit less than half of 57 minutes. Mike will therefore run a little bit less than 
laps, which is about 
.
~UltimateDL
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. 
