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?
 
  
<math>\mathrm{(A) \ } 171\qquad\mathrm{(B) \ } 173\qquad\mathrm{(C) \ } 182\qquad\mathrm{(D) \ } 188\qquad\mathrm{(E) \ } 210\qquad</math>
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{{image}}
== Solution ==
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== See Also ==
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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?
*[[2006 AMC 10A Problems]]
 
  
*[[2006 AMC 10A Problems/Problem 13|Previous Problem]]
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<math> \mathrm{(A) \ } 171\qquad \mathrm{(B) \ } 173\qquad \mathrm{(C) \ } 182\qquad \mathrm{(D) \ } 188</math><math>\mathrm{(E) \ }  210</math>
  
*[[2006 AMC 10A Problems/Problem 15|Next Problem]]
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== Solution ==
  
[[Category:Introductory Algebra Problems]]
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== See also ==
[[Category:Introductory Geometry Problems]]
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* [[2006 AMC 12A Problems]]
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*[[2006 AMC 12A Problems/Problem 11|Previous Problem]]
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*[[2006 AMC 12A Problems/Problem 13|Next Problem]]

Revision as of 17:52, 5 November 2006

Problem


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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?

$\mathrm{(A) \ } 171\qquad \mathrm{(B) \ } 173\qquad \mathrm{(C) \ } 182\qquad \mathrm{(D) \ } 188$$\mathrm{(E) \ }  210$

Solution

See also