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2004 AMC 10A Problem 2

Below is the professionally curated solution for Problem 2 of the 2004 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

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Concepts:custom operationfraction

Difficulty rating: 980

2.

For any three real numbers a,a, b,b, and c,c, with b≠c,b \neq c, the operation ⋄\diamond is defined by ⋄(a,b,c)=ab−c.\diamond(a, b, c) = \dfrac{a}{b - c}. What is ⋄(⋄(1,2,3),⋄(2,3,1),⋄(3,1,2))?\diamond(\diamond(1, 2, 3), \diamond(2, 3, 1), \diamond(3, 1, 2))?

−12-\dfrac{1}{2}

−14-\dfrac{1}{4}

00

14\dfrac{1}{4}

12\dfrac{1}{2}

Solution:

The inner values are ⋄(1,2,3)=12−3=−1,⋄(2,3,1)=23−1=1,⋄(3,1,2)=31−2=−3. \diamond(1,2,3) = \dfrac{1}{2-3} = -1, \quad \diamond(2,3,1) = \dfrac{2}{3-1} = 1, \quad \diamond(3,1,2) = \dfrac{3}{1-2} = -3.

Therefore ⋄(−1,1,−3)=−11−(−3)=−14. \diamond(-1, 1, -3) = \dfrac{-1}{1 - (-3)} = -\dfrac{1}{4}.

Thus, the correct answer is B.

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