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"

(→‎Solution 3)
(→‎Problem 22)
 
(29 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 
==Problem 22==
 
==Problem 22==
 +
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.
+
<asy>
 +
/* AMC8 P22 2024, revised by Teacher David */
 +
size(150);
  
−
(A) <math>300</math>   (B) <math>600</math>  (C) <math>1200</math>  (D) <math>1500</math>  (E) <math>1800</math>
+
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>
 +
 
 +
<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==
Line 15: Line 34:
  
 
==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 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
+
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.  
  
−
==Solution 3 (but kind of different?)==
+
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
  
−
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 = 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 <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 by Power Solve==
 
−
https://www.youtube.com/watch?v=mGsl2YZWJVU
 
  
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
Line 28: Line 54:
  
 
~Math-X
 
~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==
 
==Video Solution 2 by OmegaLearn.org==
Line 45: Line 83:
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/ktzijuZtDas&t=2679
 
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==
 
==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 20:35, 19 November 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.

[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]

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

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 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=235097"