Difference between revisions of "2022 AMC 10A Problems/Problem 2"
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− | == Solution 4 ( | + | == Solution 4 (Quick Estimate) == |
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>. | 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>. | ||
Revision as of 19:39, 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 where
is the number of laps he can complete in
minutes.
If you cross multiply,
.
So, .
~Shiloh000
Solution 4 (Quick Estimate)
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.