Difference between revisions of "2022 AMC 10A Problems/Problem 2"
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== Solution 3== | == Solution 3== | ||
− | Mike's rate is <cmath>\frac{15}{57}=\frac{x}{27},</cmath> where <math>x</math> is | + | Mike's rate is <cmath>\frac{15}{57}=\frac{x}{27},</cmath> where <math>x</math> is the number of laps he can complete in <math>27</math> minutes. |
If you cross multiply, <math>57x = 405</math>. | If you cross multiply, <math>57x = 405</math>. | ||
− | So, <math>x = \frac{405}{57} \approx \boxed{\textbf{(B) }7}</math> | + | So, <math>x = \frac{405}{57} \approx \boxed{\textbf{(B) }7}</math>. |
~Shiloh000 | ~Shiloh000 | ||
+ | |||
+ | == Solution 4 (Quick Estimate) == | ||
+ | Note that <math>27</math> minutes is a little bit less than half of <math>57</math> 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 | ||
+ | |||
+ | == Solution 5 (Approximation)== | ||
+ | Note that <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} = \boxed{\textbf{(B) }7}</math>. | ||
+ | |||
+ | ~TheGoldenRetriever | ||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== | ||
Line 30: | Line 40: | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/Zkmy3zQRzaQ | ||
== See Also == | == See Also == |
Latest revision as of 12:10, 25 December 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)
Note that 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
.
~TheGoldenRetriever
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
Video Solution 2
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.