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

Revision as of 14:02, 15 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 $\frac{15}{57}=\frac{x}{27}$ laps per minute where x is how many laps he can complete in $27$ minutes.

If you cross multiply, $57x$ = $405$ .

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

~Shiloh000

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

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

@@ Line 17: / Line 17: @@
 ~MrThinker
+== Solution 3==
+Mike's rate is <math>\frac{15}{57}=\frac{x}{27}</math> laps per minute where x is how many laps he can complete in <math>27</math> minutes.
+If you cross multiply, <math>57x</math> = <math>405</math>.
+So, x is roughly <math>\frac{405}{57} \approx \boxed{\textbf{(B) }7}</math> laps.
+~Shiloh000
 ==Video Solution 1 (Quick and Easy)==

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