2005 AMC 10A Problem 2 — Solution | LIVE by Po-Shen Loh

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2005 AMC 10A Problem 2

Below is the professionally curated solution for Problem 2 of the 2005 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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Concepts:custom operation fraction order of operations

Difficulty rating: 960

2.

For each pair of real numbers $a \neq b,$ define the operation $\star$ as $(a \star b) = \frac{a+b}{a-b}.$

What is the value of $((1 \star 2) \star 3)?$

$-\dfrac{2}{3}$

$-\dfrac{1}{5}$

$0$

$\dfrac{1}{2}$

This value is not defined.

Solution:

First $(1 \star 2) = \dfrac{1+2}{1-2} = \dfrac{3}{-1} = -3.$ Then $(-3 \star 3) = \dfrac{-3+3}{-3-3} = \dfrac{0}{-6} = 0.$

Thus, the correct answer is C.

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