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2006 AMC 12A Problems/Problem 12

Revision as of 20:28, 14 February 2007 by Klactoveesedstene (talk | contribs) (→‎See Also)

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?

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

Solution

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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)

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