AMC 8 · 2024 · #22

Grade 7 geometry-2d

Archetype Arithmetic Expression Decomposition

area-circlesrate dimensional-analysisarea-difference ↑ Prerequisites: area-circlesfraction-decimal-conversion

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

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.

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

Pick an answer.

(A)

300

(B)

600

(C)

1200

(D)

1500

(E)

1800

View mode:

Toolkit + CCSS Solution

Understand

Restated: A roll of tape looks like a flat ring (an **annulus**) when you look at it end-on: the outer disk has diameter $4$ in and there's a hole in the middle of diameter $2$ in. The tape itself is $0.015$ in thick. If you unrolled the whole thing into one long flat strip, about how long would it be (rounded to the nearest $100$ inches)?

Givens: Outer diameter of the roll: $4$ in (so outer radius $R = 2$ in); Inner ring (hole) diameter: $2$ in (so inner radius $r = 1$ in); Thickness of the tape: $t = 0.015$ in; Answer choices: (A) 300, (B) 600, (C) 1200, (D) 1500, (E) 1800

Unknowns: The approximate length $L$ of the unrolled tape, in inches, rounded to the nearest $100$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #10 Create a Physical Representation, #7 Identify Subproblems, #3 Eliminate Possibilities

The key trick is to **draw two pictures side by side** (Tool #1): the ring-shaped cross-section of the roll, and the same tape unrolled into a long thin rectangle. Imagining the unrolling physically (Tool #10) makes it obvious that the same amount of tape-material is just rearranged — the cross-sectional area doesn't change. Then we break the work into clean subproblems (Tool #7): (a) area of the ring, (b) length from area $=$ length $\times$ thickness, (c) round. Finally we use Tool #3 to match the rounded value to a choice. This avoids reaching for algebra (Tool #13); the only formula we truly need is **area of a circle**.

Execute — Answer: B

#10 Create a Physical Representation 3.MD.C.5 Step 1

Picture the tape **rolled up** as a flat ring, and the same tape **unrolled** as a long thin rectangle whose width is the tape's thickness $t = 0.015$ in and whose length is the unknown $L$.
The amount of tape-stuff doesn't change when you unroll it, so the area of the ring (looking at the roll end-on) must equal the area of the rectangle.
That single observation is the whole strategy: $\text{Area of ring} = L \times t$.

$$\text{Area}_{\text{ring}} = L \times t$$

💡 Area is the amount of flat space a shape covers; unrolling rearranges the tape but doesn't change how much flat space it covers — exactly the Grade 3 idea that area is an attribute of a plane figure.

#1 Draw a Diagram 3.OA.A.2 Step 2

Read the radii off the picture.
Diameters are twice the radii, so divide each diameter by $2$: outer radius $R = 4 \div 2 = 2$ in, inner radius $r = 2 \div 2 = 1$ in.
Label them on your diagram next to the two circles.

$$R = \tfrac{4}{2} = 2 \text{ in},\quad r = \tfrac{2}{2} = 1 \text{ in}$$

💡 Diameter $\div 2 =$ radius is a simple Grade 3 division-into-equal-parts move.

#7 Identify Subproblems 7.G.B.4 Step 3

Find the **area of the ring** by subtracting: big disk area minus small (hole) disk area.
Use the circle area formula $\pi r^2$.
Big disk: $\pi \cdot 2^2 = 4\pi$.
Small disk: $\pi \cdot 1^2 = \pi$.
Ring area $= 4\pi - \pi = 3\pi$ square inches.

$$\text{Area}_{\text{ring}} = \pi R^2 - \pi r^2 = \pi(2^2) - \pi(1^2) = 4\pi - \pi = 3\pi$$

💡 Knowing the area-of-a-circle formula $\pi r^2$ is exactly the Grade 7 standard for circles; subtracting a smaller disk from a bigger one is straightforward composition.

#7 Identify Subproblems 6.NS.B.3 Step 4

Plug into $\text{Area}_{\text{ring}} = L \times t$ and solve for $L$.
We have $3\pi = L \times 0.015$, so $L = \dfrac{3\pi}{0.015}$.
Approximate with $\pi \approx 3.14$: $3\pi \approx 9.42$, and $\dfrac{9.42}{0.015} = \dfrac{9420}{15} = 628$ inches.
(Or, exactly: $0.015 = \tfrac{3}{200}$, so $L = 3\pi \div \tfrac{3}{200} = 200\pi \approx 628$.)

$$L = \dfrac{3\pi}{0.015} = \dfrac{9.42}{0.015} = 628 \text{ in}$$

💡 Dividing $9.42$ by $0.015$ (or, equivalently, $9420 \div 15$) is exactly the Grade 6 fluent multi-digit decimal division skill.

#3 Eliminate Possibilities 3.NBT.A.1 Step 5

The problem says round to the nearest $100$.
The hundreds digit of $628$ is $6$ and the tens digit is $2$, which is less than $5$, so we round **down** to $600$.
Match against the choices: $600$ is exactly choice (B).

$$628 \approx 600 \;\Rightarrow\; \textbf{(B)}$$

💡 Rounding a three-digit whole number to the nearest hundred is the Grade 3 rounding standard.

[1] #10 3.MD.C.5 Picture the tape **rolled up** as a flat ring, and the same tape **unrolled** as

[2] #1 3.OA.A.2 Read the radii off the picture. Diameters are twice the radii, so divide each di

[3] #7 7.G.B.4 Find the **area of the ring** by subtracting: big disk area minus small (hole) d

[4] #7 6.NS.B.3 Plug into $\text{Area}_{\text{ring}} = L \times t$ and solve for $L$. We have $3

[5] #3 3.NBT.A.1 The problem says round to the nearest $100$. The hundreds digit of $628$ is $6$

Review

Reasonableness: Sanity-check the size. The roll has outer radius $2$ in, so its outer circumference is only $2\pi \approx 6.3$ in — yet we got a length of about $628$ in (over $52$ ft!). Does $628$ make sense? Yes: the tape is extremely thin ($0.015$ in), and roughly $\tfrac{2 - 1}{0.015} \approx 67$ layers stack between $r = 1$ and $R = 2$. Each layer is on average about $2\pi \cdot 1.5 \approx 9.4$ in around, and $67 \times 9.4 \approx 630$ in — matching $628$ almost exactly. Good. Also $200\pi$ being our exact answer is a clean number, which is reassuring.

Alternative: Instead of equating areas, you could use the layer-counting estimate from the reasonableness check directly: number of layers $\approx \tfrac{R - r}{t} = \tfrac{1}{0.015} \approx 67$, and average layer length $\approx 2\pi \cdot \tfrac{R + r}{2} = 2\pi \cdot 1.5 = 3\pi$. Total $\approx 67 \times 3\pi \approx 200\pi \approx 628$. This uses Tool #5 (Pattern) and Tool #9 (Easier Problem — average circumference), and lands on the same $\approx 600$ in.

CCSS standards used (min grade 7)

3.MD.C.5 Recognize area as an attribute of plane figures and understand concepts (Recognizing that unrolling the tape rearranges but does not change the cross-sectional area, so ring area $= L \times t$.)
3.OA.A.2 Interpret whole-number quotients of whole numbers (Halving each diameter to get the radii: $R = 4 \div 2 = 2$ and $r = 2 \div 2 = 1$.)
3.NBT.A.1 Round whole numbers to the nearest 10 or 100 (Rounding $628$ to the nearest $100$ to get $600$.)
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals (Computing $L = 9.42 \div 0.015 = 628$ to evaluate $3\pi / 0.015$.)
7.G.B.4 Know the formulas for area and circumference of a circle (Using $\pi r^2$ to find the area of the outer disk and the inner disk, then subtracting to get the ring area $3\pi$.)

⭐ This AMC 8 problem only needs Grade 7 circle-area $\pi r^2$ you already know — plus a clever picture that unrolls the tape into a thin rectangle!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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