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

Revision as of 13:05, 8 February 2024

1 Problem 22
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 3 (but kind of different?)
6 Video Solution by Power Solve
7 Video Solution 1 by Math-X (First understand the problem!!!)
8 Video Solution 2 by OmegaLearn.org
9 Video Solution (Arithmetic Series)3 by SpreadTheMathlove Using
10 Video Solution by NiuniuMaths (Easy to understand!)
11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
12 Video Solution by Interstigation
13 See Also

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.

(A) $300$ (B) $600$ (C) $1200$ (D) $1500$ (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 $X = \boxed {600}$ .

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/BaE00H2SHQM?si=5YWwuZ_961azySZ-&t=6686

~Math-X

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

@@ Line 19: / Line 19: @@
 ==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 <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 Y's give us an equation for X, and if we approximate <math>\pi</math> as 3, then X = <math>\boxed {600}</math>.
+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>.
 ==Video Solution by Power Solve==

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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