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

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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 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.  
 
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.  
  
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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 <math>\pi</math> to be 3, we get the final answer to be <math>200 \cdot 3 = \textbf {(B) } 600</math>. -IwOwOwl253
+
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>.  
 +
-IwOwOwl253
  
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==Solution 3 (but kind of different?)==
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==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!
 
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==
 
==Solution 4==
 
If you cannot notice that the average diameter is <math>3</math>,
 
If you cannot notice that the average diameter is <math>3</math>,
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1. you can still solve this problem by the following method.  
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you can still solve this problem by the following method.  
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2. You stoobid
 
  
 
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>.
 
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==
+
==Video Solution (A Clever Explanation You’ll Get Instantly)==
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https://youtu.be/NTJM_U-GhlM
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https://youtu.be/5ZIFnqymdDQ?si=MFF7Oc2wqpOhm2UU&t=3250
  
−
Please like and sub
+
~hsnacademy
  
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
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~Math-X
 
~Math-X
 +
 +
== Video solution==
 +
https://youtu.be/NTJM_U-GhlM
 +
 +
Please like and sub
 +
 
==Video Solution by Power Solve==
 
==Video Solution by Power Solve==
 
https://www.youtube.com/watch?v=mGsl2YZWJVU
 
https://www.youtube.com/watch?v=mGsl2YZWJVU

Latest revision as of 09:14, 6 September 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 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 $\pi(\frac{4}{2})^2 -\pi(\frac{2}{2})^2 = 3\pi$, and we divide that by $0.015$ to get $200\pi$. Approximating $\pi$ to be 3, we get the final answer to be $200 \cdot 3 = \textbf {(B) } 600$. -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 $X$ (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 (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

https://youtu.be/NTJM_U-GhlM

Please like and sub

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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