Difference between revisions of "2006 AMC 12A Problems/Problem 12"
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== Problem == | == Problem == | ||
| − | + | A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outisde diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring? | |
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| − | A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an | ||
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| + | <math>\mathrm{(A) \ } 171\qquad\mathrm{(B) \ } 173\qquad\mathrm{(C) \ } 182\qquad\mathrm{(D) \ } 188\qquad\mathrm{(E) \ } 210\qquad</math> | ||
== Solution == | == Solution == | ||
| − | + | == See Also == | |
| + | *[[2006 AMC 10A Problems]] | ||
| − | + | *[[2006 AMC 10A Problems/Problem 13|Previous Problem]] | |
| − | * [[2006 AMC | ||
| − | + | *[[2006 AMC 10A Problems/Problem 15|Next Problem]] | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | Solution | ||
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| + | Add the inner diameters, which go from 18 down to 1. Then add 2 more for the thickness of the top and bottom rings. | ||
| + | (18)(19)/2 + 2 = 173 | ||
| + | |||
| + | (B) | ||
Revision as of 20:28, 14 February 2007
Problem
A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outisde diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
Solution
See Also
Solution
Add the inner diameters, which go from 18 down to 1. Then add 2 more for the thickness of the top and bottom rings. (18)(19)/2 + 2 = 173
(B)
