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Difference between revisions of "2024 AMC 8 Problems/Problem 22"

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==Problem==
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==Problem 22==
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imagine trying to cheat on the amc8 of all things
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  hi agine
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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.
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<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==
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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>.
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-ILoveMath31415926535
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==Solution 2==
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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>.
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==Solution 3==
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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 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 <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 pi to be 3, we get the final answer to be <math>200 \cdot 3 = (B) 600</math>. -IwOwOwl253
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==Solution 3 (but kind of different?)==
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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!
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==Solution 4==
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If you cannot notice that the average diameter is <math>3</math>, you can still solve this problem by the following method.
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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>.
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==Video Solution 1 by Math-X (First understand the problem!!!)==
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https://youtu.be/BaE00H2SHQM?si=5YWwuZ_961azySZ-&t=6686
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~Math-X
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==Video Solution by Power Solve==
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https://www.youtube.com/watch?v=mGsl2YZWJVU
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==Video Solution 2 by OmegaLearn.org==
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https://youtu.be/k1yAO0pZw-c
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==Video Solution (Arithmetic Series)3 by SpreadTheMathlove Using ==
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https://www.youtube.com/watch?v=kv_id-MgtgY
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== Video Solution by NiuniuMaths (Easy to understand!) ==
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https://www.youtube.com/watch?v=uAHP_LPUcwQ
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~NiuniuMaths
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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https://www.youtube.com/watch?v=bldjKBbhvkE
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==Video Solution by Interstigation==
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https://youtu.be/ktzijuZtDas&t=2679
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==See Also==
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{{AMC8 box|year=2024|num-b=21|num-a=23}}
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{{MAA Notice}}

Revision as of 05:39, 4 March 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.

$\textbf{(A) } 300\qquad\textbf{(B) } 600\qquad\textbf{(C) } 1200\qquad\textbf{(D) } 1500\qquad\textbf{(E) } 1800$

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

Solution 3 (but kind of different?)

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 $X$ (this is what the problem wants!) and the width is $Y$. Then, the volume is, of course, $0.015 \cdot X \cdot Y.$ 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, $3\pi \cdot Y.$ Now, since the volume always stays the same, we know that $3\pi \cdot Y = 0.015 \cdot X \cdot Y.$ Cancelling the $Y$'s give us an equation for $X$, and if we approximate $\pi$ as $3$, then $X = \boxed {600}$. Yay!


Solution 4

If you cannot notice that the average diameter is $3$, you can still solve this problem by the following method.

The same with solution 1, we have $\frac{1000}{0.015}$ layers of tape. If we consider every layers with the diameter $2$, the length should be $\frac{1000}{0.015}2\pi\approx 400$. If the diameter is seem as $4$, the length should be $800$. So, the length is between $400$ and $800$, the only possible answer is $\boxed{600}$.

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

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

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

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