2024 AMC 8 Problems/Problem 22
Contents
- 1 Problem 22
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 3 (kind of different?, but fun!)
- 6 Solution 4
- 7 Video Solution (A Clever Explanation You’ll Get Instantly)
- 8 Video Solution 1 by Math-X (First understand the problem!!!)
- 9 Video solution
- 10 Video Solution by Power Solve
- 11 Video Solution 2 by OmegaLearn.org
- 12 Video Solution (Arithmetic Series)3 by SpreadTheMathlove Using
- 13 Video Solution by NiuniuMaths (Easy to understand!)
- 14 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 15 Video Solution by Interstigation
- 16 Video Solution by Dr. David
- 17 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.
Solution 1
The roll of tape is 66 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 of the tape as a really thin rectangle that has a width of inches. The side of the tape is wrapped into an annulus(The shaded region between 2 circles with the same center), meaning the area of the shaded region is equal to the area of the really thin rectangle.
The area of the shaded region is , and we divide that by to get . Approximating to be 3, we get the final answer to be . -IwOwOwl253
Solution 3 (kind of different?, but fun!)
The volume of the tape is always the same, but we can either calculate it when the tape is unrolled as a really long, thin rectangular prism, or we can calculate it as a cylinder with a hole cut out of it. When we calculate it as a long rectangular prism, we can say that the length is (this is what the problem wants!) and the width is . Then, the volume is, of course, Now, notice that the "width" of our rectangular prism is also the "height" of our cylinder with a hole cut out of it. Then, we can calculate the volume as base times height, or in this case, Now, since the volume always stays the same, we know that Cancelling the 's give us an equation for , and if we approximate as , then . Yay!
Solution 4
If you cannot notice that the average diameter is , you can still solve this problem by the following method.
The same with solution 1, we have layers of tape. If we consider every layers with the diameter , the length should be . If the diameter is seem as , the length should be . So, the length is between and , the only possible answer is .
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=MFF7Oc2wqpOhm2UU&t=3250
~hsnacademy
Video Solution 1 by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=5YWwuZ_961azySZ-&t=6686
~Math-X
Video solution
Please like and sub
Video Solution by Power Solve
https://www.youtube.com/watch?v=mGsl2YZWJVU
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 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
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=2679
Video Solution by Dr. David
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.