Difference between revisions of "2006 AMC 12A Problems/Problem 12"

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* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
  
{{AMC box|year=2006|n=12A|num-b=11|num-a=13}}
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 18:19, 2 February 2007

Problem


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A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside 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

The sum of the consecutively increasing integers from 3 to 20 is $\frac{1}{2}(18)(3+20) = 207$. However, the 17 intersections between the rings must also be subtracted, so we get $207 - 2(17) = 173 \Rightarrow B$.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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
All AMC 12 Problems and Solutions