Art of Problem Solving
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2024 AMC 8 Problems/Problem 22: Difference between revisions

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==Problem 22==
==Problem 22==
A roll of tape is <math>4</math> inches in diameter and is wrapped around a ring that is <math>2</math> inches in diameter. A cross section of the tape is shown in the figure below. The tape is <math>0.015</math> inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest <math>100</math> inches.
(A) <math>300</math>  (B) <math>600</math>  (C) <math>1200</math>  (D) <math>1500</math>  (E) <math>1800</math>


==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>

Revision as of 16:28, 25 January 2024

Problem 22

A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.

(A) $300$ (B) $600$ (C) $1200$ (D) $1500$ (E) $1800$

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 $600.$