2024 AMC 8 Problems/Problem 22
Contents
[hide]- 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 CosineMethod [🔥Fast and Easy🔥]
- 10 See Also
Problem 22
A roll of tape is inches in diameter and is wrapped around a ring that is inches in diameter. A cross section of the tape is shown in the figure below. The tape is inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest inches.
(A) (B) (C) (D) (E)
Solution 1
The roll of tape is layers thick. In order to find the total length, we have to find the average of each concentric circle and multiply it by . Since the diameter of the small circle is inches and the diameter of the large one is inches, the "middle value" is . Therefore, the average circumference is . Multiplying gives .
-ILoveMath31415926535
Solution 2
There are about "full circles" of tape, and with average circumference of which means the answer is
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 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 0.015200\pi200 \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
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 | |
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All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.