Difference between revisions of "2024 AMC 8 Problems/Problem 22"
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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 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>\textbf{(A) } 300\qquad\textbf{(B) } 600\qquad\textbf{(C) } 1200\qquad\textbf{(D) } 1500\qquad\textbf{(E) } 1800</math> |
==Solution 1== | ==Solution 1== | ||
− | The roll of tape is <math>1/0.015 | + | The roll of tape is <math>1/0.015=</math>66 layers thick. In order to find the total length, we have to find the average of each concentric circle and multiply it by <math>66</math>. Since the diameter of the small circle is <math>2</math> inches and the diameter of the large one is <math>4</math> inches, the "middle value" is <math>3</math>. Therefore, the average circumference is <math>3\pi</math>. Multiplying <math>3\pi \cdot 66</math> gives <math>(B) \boxed{600}</math>. |
-ILoveMath31415926535 | -ILoveMath31415926535 | ||
==Solution 2== | ==Solution 2== | ||
− | 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} | + | 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} \cdot 3\pi=200\pi, </math> which means the answer is <math>\boxed{600}</math>. |
==Solution 3== | ==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 <math>0.015</math> inches. The side of the tape is wrapped into | + | 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 <math>0.015</math> 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 <math>\pi(\frac{4}{2})^2 -\pi(\frac{2}{2})^2 = 3\pi</math>, and we divide that by <math>0.015</math> to get <math>200\pi</math>. Approximating <math>\pi</math> to be 3, we get the final answer to be <math>200 \cdot 3 = \textbf {(B) } 600</math>. |
− | https:// | + | -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 <math>X</math> (this is what the problem wants!) and the width is <math>Y</math>. Then, the volume is, of course, <math>0.015 \cdot X \cdot Y.</math> 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, <math>3\pi \cdot Y.</math> Now, since the volume always stays the same, we know that <math>3\pi \cdot Y = 0.015 \cdot X \cdot Y.</math> Cancelling the <math>Y</math>'s give us an equation for <math>X</math>, and if we approximate <math>\pi</math> as <math>3</math>, then <math>X = \boxed {600}</math>. Yay! | ||
+ | |||
+ | ==Solution 4== | ||
+ | If you cannot notice that the average diameter is <math>3</math>, | ||
+ | you can still solve this problem by the following method. | ||
+ | |||
+ | The same with solution 1, we have <math>\frac{1000}{0.015}</math> layers of tape. If we consider every layers with the diameter <math>2</math>, the length should be <math>\frac{1000}{0.015}2\pi\approx 400</math>. If the diameter is seem as <math>4</math>, the length should be <math>800</math>. So, the length is between <math>400</math> and <math>800</math>, the only possible answer is <math>\boxed{600}</math>. | ||
+ | |||
+ | == Video solution== | ||
+ | https://youtu.be/NTJM_U-GhlM | ||
+ | |||
+ | Please like and sub | ||
==Video Solution 1 by Math-X (First understand the problem!!!)== | ==Video Solution 1 by Math-X (First understand the problem!!!)== | ||
− | https://youtu.be/ | + | https://youtu.be/BaE00H2SHQM?si=5YWwuZ_961azySZ-&t=6686 |
~Math-X | ~Math-X | ||
+ | ==Video Solution by Power Solve== | ||
+ | https://www.youtube.com/watch?v=mGsl2YZWJVU | ||
==Video Solution 2 by OmegaLearn.org== | ==Video Solution 2 by OmegaLearn.org== | ||
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==Video Solution (Arithmetic Series)3 by SpreadTheMathlove Using == | ==Video Solution (Arithmetic Series)3 by SpreadTheMathlove Using == | ||
https://www.youtube.com/watch?v=kv_id-MgtgY | 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🔥]== | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
https://www.youtube.com/watch?v=bldjKBbhvkE | https://www.youtube.com/watch?v=bldjKBbhvkE | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/ktzijuZtDas&t=2679 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2024|num-b=21|num-a=23}} | {{AMC8 box|year=2024|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:56, 9 August 2024
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
- 8 Video Solution 1 by Math-X (First understand the problem!!!)
- 9 Video Solution by Power Solve
- 10 Video Solution 2 by OmegaLearn.org
- 11 Video Solution (Arithmetic Series)3 by SpreadTheMathlove Using
- 12 Video Solution by NiuniuMaths (Easy to understand!)
- 13 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 14 Video Solution by Interstigation
- 15 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
Please like and sub
Video Solution 1 by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=5YWwuZ_961azySZ-&t=6686
~Math-X
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
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.