Difference between revisions of "1987 AHSME Problems/Problem 13"
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(Added a solution with explanation) |
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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 == | ||
| + | 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 
cm wide is made into a roll for cash registers by wrapping it 
times around a cardboard tube of diameter 
cm,
forming a roll 
cm in diameter. Approximate the length of the paper in meters.
(Pretend the paper forms concentric circles with diameters evenly spaced from cm to cm.)
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 
times the sum of the diameters. Now the, the diameters form an arithmetic series with first term , last term , and terms in total, so using the formula 
, the sum is 
, so the length is 
centimetres, or 
metres, which is answer 
.
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
