2022 AMC 10A Problems/Problem 2: Difference between revisions

Revision as of 18:39, 21 November 2022

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

Video Solution 1 (Quick and Easy)

https://youtu.be/tu4rE1nqY9g

~Education, the Study of Everything

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

@@ Line 26: / Line 26: @@
 ~Shiloh000
-== Solution 4 (Reasonable Guess) ==
+== 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>.

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