2024 AMC 8 Problems/Problem 22

1 Problem 22
2 Solution 1
3 Solution 2
4 Solution 3
5 Video Solution by Power Solve
6 Video Solution 1 by Math-X (First understand the problem!!!)
7 Video Solution 2 by OmegaLearn.org
8 Video Solution (Arithmetic Series)3 by SpreadTheMathlove Using
9 Video Solution by NiuniuMaths (Easy to understand!)
10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
11 See Also

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 $3\pi \cdot 66$ 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 $\dfrac{4+2}{2}\pi=3\pi.$ $\dfrac{200}{3} \cdot 3\pi=200\pi,$ which means the answer is $\boxed{600}$ .

Solution 3

We can figure out the length of the tape by considering the side of the tape as a really thin rectangle that has a width of $0.015$ inches. The side of the tape is wrapped into a annulus(The shaded region between 2 circles with the same center), meaning the area of the circle is equal to the area of the rectangle. The area of the annulus is $\pi(\frac{4}{2})^2 -\pi(\frac{2}{2})^2 = 3\pi$ , and we divide that by $0.015$ to get $200\pi$ . Approximating pi to be 3, we get the final answer to be $200 \cdot 3 = (B) 600$ . -IwOwOwl253

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 NiuniuMaths (Easy to understand!)

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

~NiuniuMaths

Video Solution by CosineMethod [🔥Fast and Easy🔥]

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

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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