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

Latest revision as of 10:58, 29 October 2024

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

Video Solution 3

https://www.youtube.com/watch?v=bKtyNh5hPWI

~Math4All999

@@ Line 4: / Line 4: @@
 <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.
@@ Line 11: / Line 11: @@
 ~MRENTHUSIASM
+== Solution 2==
+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.
+~MrThinker
+== Solution 3==
+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>.
+So, <math>x = \frac{405}{57} \approx \boxed{\textbf{(B) }7}</math>.
+~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)==
+https://youtu.be/tu4rE1nqY9g
+~Education, the Study of Everything
+==Video Solution 2==
+https://youtu.be/Zkmy3zQRzaQ
+==Video Solution 3==
+https://www.youtube.com/watch?v=bKtyNh5hPWI
+~Math4All999
 == See Also ==

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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