Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
(→‎Solution 3 (kind of different?, but fun!))
 
(2 intermediate revisions by the same user not shown)
Line 34: Line 34:
 
The roll of tape is <math>1/0.015=66</math> 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" (or mean) is <math>3</math>. Therefore, the average circumference is <math>3\pi</math>. Multiplying <math>3\pi \cdot 66</math> gives approximately <math>(B) \boxed{600}</math>.
 
The roll of tape is <math>1/0.015=66</math> 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" (or mean) is <math>3</math>. Therefore, the average circumference is <math>3\pi</math>. Multiplying <math>3\pi \cdot 66</math> gives approximately <math>(B) \boxed{600}</math>.
 
- Ilovemath3141592653589
 
- Ilovemath3141592653589
−
 
−
-Minor edit by angieeverfree
 
  
 
==Solution 2==
 
==Solution 2==
Line 43: Line 41:
 
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.  
  
−
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>.  
+
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
 
  
 
==Solution 3 (kind of different?, but fun!)==
 
==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>.
  
 
==Solution 4==
 
==Solution 4==

Latest revision as of 05:38, 11 January 2025

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.

[asy] /* AMC8 P22 2024, revised by Teacher David */ size(150); pair o = (0,0); real r1 = 1; real r2 = 2; filldraw(circle(o, r2), mediumgray, linewidth(1pt)); filldraw(circle(o, r1), white, linewidth(1pt)); draw((-2,-2.6)--(-2,-2.4)); draw((2,-2.6)--(2,-2.4)); draw((-2,-2.5)--(2,-2.5), L=Label("4 in.")); draw((-1,0)--(1,0), L=Label("2 in.", align=(0,1)), arrow=Arrows()); draw((2,0)--(2,-1.3), linewidth(1pt)); [/asy]

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

Official Video Solution (Real-world visualization method- 2 minutes!)

https://www.youtube.com/watch?v=UL_xlVUKcw8&t=7s ~TheMathGeek


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" (or mean) is $3$. Therefore, the average circumference is $3\pi$. Multiplying $3\pi \cdot 66$ gives approximately $(B) \boxed{600}$. - Ilovemath3141592653589

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 $200 \cdot 3 = \textbf {(B) } 600$.

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 $X = \boxed {600}$.

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 by Central Valley Math Circle (Goes through full thought process)

https://youtu.be/cJUWOQ-tSnE

~mr_mathman

Video Solution 1 by Math-X (First understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=5YWwuZ_961azySZ-&t=6686

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=MFF7Oc2wqpOhm2UU&t=3250 ~hsnacademy

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

Video Solution by Dr. David

https://youtu.be/vnG8JYpuaJM

Video Solution by WhyMath

https://youtu.be/k2FNc1sf-sE

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_22&oldid=239668"