1987 AHSME Problems/Problem 13

Problem

A long piece of paper $5$ cm wide is made into a roll for cash registers by wrapping it $600$ times around a cardboard tube of diameter $2$ cm, forming a roll $10$ cm in diameter. Approximate the length of the paper in meters. (Pretend the paper forms $600$ concentric circles with diameters evenly spaced from $2$ cm to $10$ cm.)

$\textbf{(A)}\ 36\pi \qquad \textbf{(B)}\ 45\pi \qquad \textbf{(C)}\ 60\pi \qquad \textbf{(D)}\ 72\pi \qquad \textbf{(E)}\ 90\pi$

Solution 1

Notice (by imagining unfolding the roll), that the length of the paper is equal to the sum of the circumferences of the concentric circles, which is $\pi$ times the sum of the diameters. Now the, the diameters form an arithmetic series with first term $2$ , last term $10$ , and $600$ terms in total, so using the formula $\frac{1}{2}n(a+l)$ , the sum is $300 \times 12 = 3600$ , so the length is $3600\pi$ centimetres, or $36\pi$ metres, which is answer $\boxed{\text{A}}$ .

Solution 2

When the tape is unrolled, its cross-sectional area must remain constant. The tape's cross-sectional area must be $\pi\cdot 5^2 - \pi \cdot 1^2 = 24\pi$ square centimeters. Since the roll is $5 - 1 = 4$ cm thick and has $600$ coils, the tape must be $\frac{4}{600} = \frac{1}{150}$ cm thick. The unrolled tape forms a rectangular prism, so its length must be $\frac{24\pi}{\frac{1}{150}} = 3600\pi$ cm, or $\boxed{(\mathbf{A})\ 36\pi}$ m.

-j314andrews

1987 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1987 AHSME Problems/Problem 13

Contents

Problem

Solution 1

Solution 2

See also