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Difference between revisions of "2024 AMC 8 Problems/Problem 22"
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==Solution 3==
 
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We can figure out the length of the tape by considering the side as a really thin rectangle that has a width of 0.015.
  
 
==Video Solution by Power Solve==
 
==Video Solution by Power Solve==

Revision as of 22:25, 26 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 1

The roll of tape is $1/0.015~66$ layers thick. In order to find the total length, we have to find the average of each concentric circle and multiply it by $66$. Since the diameter of the small circle is $2$ inches and the diameter of the large one is $4$ inches, the "middle value" is $3$. Therefore, the average circumference is $3\pi$. Multiplying $3\pi*66$ gives $(B) \boxed{600}$.

-ILoveMath31415926535

Solution 2

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 $600.$

Solution 3

We can figure out the length of the tape by considering the side as a really thin rectangle that has a width of 0.015.

Video Solution by Power Solve

https://www.youtube.com/watch?v=mGsl2YZWJVU

Video Solution 1 by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=Ngh2w5-AAuP38GZk&t=34

~Math-X

Video Solution 2 by OmegaLearn.org

https://youtu.be/k1yAO0pZw-c

Video Solution (Arithmetic Series)3 by SpreadTheMathlove Using

https://www.youtube.com/watch?v=kv_id-MgtgY

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=bldjKBbhvkE

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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