# 2022 AMC 10A Problems/Problem 2

## 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)

~Education, the Study of Everything

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 