AMC 8 · 2024 · #22
Grade 7 geometry-2dProblem
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.
Pick an answer.
AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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
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
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: $\text{Area of ring} = L \times t$.
💡 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.
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.
💡 Diameter $\div 2 =$ radius is a simple Grade 3 division-into-equal-parts move.
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.
💡 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.
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 $L = 3\pi \div \tfrac{3}{200} = 200\pi \approx 628$.)
💡 Dividing $9.42$ by $0.015$ (or, equivalently, $9420 \div 15$) is exactly the Grade 6 fluent multi-digit decimal division skill.
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).
💡 Rounding a three-digit whole number to the nearest hundred is the Grade 3 rounding standard.
3.MD.C.5 Picture the tape **rolled up** as a flat ring, and the same tape **unrolled** as 3.OA.A.2 Read the radii off the picture. Diameters are twice the radii, so divide each di 7.G.B.4 Find the **area of the ring** by subtracting: big disk area minus small (hole) d 6.NS.B.3 Plug into $\text{Area}_{\text{ring}} = L \times t$ and solve for $L$. We have $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.5Recognize 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.2Interpret 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.1Round whole numbers to the nearest 10 or 100 (Rounding $628$ to the nearest $100$ to get $600$.)6.NS.B.3Fluently 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.4Know 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!
⭐ 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!
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