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Difference between revisions of "2024 AMC 8 Problems/Problem 22"

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==Solution==
 
==Solution==
 
There are about <math>\dfrac{1}{0.015}=\dfrac{200}{3}</math> "full circles" of tape, and with average circumference of <math>\dfrac{4+2}{2}\pi=3\pi.</math> <math>\dfrac{200}{3}*3\pi=200\pi, </math> which means the answer is <math>600.</math>
 
There are about <math>\dfrac{1}{0.015}=\dfrac{200}{3}</math> "full circles" of tape, and with average circumference of <math>\dfrac{4+2}{2}\pi=3\pi.</math> <math>\dfrac{200}{3}*3\pi=200\pi, </math> which means the answer is <math>600.</math>
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~BS2012

Revision as of 15:56, 25 January 2024

Problem 22

Solution

There are about $\dfrac{1}{0.015}=\dfrac{200}{3}$ "full circles" of tape, and with average circumference of $\dfrac{4+2}{2}\pi=3\pi.$ $\dfrac{200}{3}*3\pi=200\pi,$ which means the answer is $600.$

~BS2012

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