Difference between revisions of "1987 AHSME Problems/Problem 13"

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\textbf{(D)}\ 72\pi \qquad
 
\textbf{(D)}\ 72\pi \qquad
 
\textbf{(E)}\ 90\pi  </math>   
 
\textbf{(E)}\ 90\pi  </math>   
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== Solution ==
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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 <math>\pi</math> times the sum of the diameters. Now the, the diameters form an arithmetic series with first term <math>2</math>, last term <math>10</math>, and <math>600</math> terms in total, so using the formula <math>\frac{1}{2}n(a+l)</math>, the sum is <math>300 \times 12 = 3600</math>, so the length is <math>3600\pi</math> centimetres, or <math>36\pi</math> metres, which is answer <math>\boxed{\text{A}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:24, 1 March 2018

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

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 $\boxed{\text{A}}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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