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Difference between revisions of "2022 AMC 10A Problems/Problem 2"

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)==
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== Solution 5 (Approximation)==
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<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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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
 
~TheGoldenRetriever
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~Education, the Study of Everything
 
~Education, the Study of Everything
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 +
==Video Solution 2==
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https://youtu.be/Zkmy3zQRzaQ
  
 
== See Also ==
 
== See Also ==

Revision as of 11:10, 25 December 2023

Problem

Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?

$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$

Solution 1

Mike's speed is $\frac{15}{57}=\frac{5}{19}$ laps per minute.

In the first $27$ minutes, he completed approximately $\frac{5}{19}\cdot27\approx\frac{1}{4}\cdot28=\boxed{\textbf{(B) } 7}$ laps.

~MRENTHUSIASM

Solution 2

Mike runs $1$ lap in $\frac{57}{15}=\frac{19}{5}$ minutes. So, in $27$ minutes, Mike ran about $\frac{27}{\frac{19}{5}} \approx \boxed{\textbf{(B) }7}$ laps.

~MrThinker

Solution 3

Mike's rate is \[\frac{15}{57}=\frac{x}{27},\] where $x$ is the number of laps he can complete in $27$ minutes. If you cross multiply, $57x = 405$.

So, $x = \frac{405}{57} \approx \boxed{\textbf{(B) }7}$.

~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 $15/2=7.5$ laps, which is about $\boxed{\textbf{(B) }7}$.

~UltimateDL

Solution 5 (Approximation)

Note that $57$ minutes is almost equal to $1$ hour. Running $15$ laps in $1$ hour is running approximately $1$ lap every $4$ minutes. This means that in $27$ minutes, Mike will run approximately $\frac{27}{4}$ laps. This is very close to $\frac{28}{4} = \boxed{\textbf{(B) }7}$.

~TheGoldenRetriever

Video Solution 1 (Quick and Easy)

https://youtu.be/tu4rE1nqY9g

~Education, the Study of Everything

Video Solution 2

https://youtu.be/Zkmy3zQRzaQ

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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