2007 AMC 10A Problems/Problem 2

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Problem

Define $a@b = ab - b^{2}$ and $a\#b = a + b - ab^{2}$ . What is $\frac {6@2}{6\#2}$ ?

$\text{(A)}\ - \frac {1}{2}\qquad \text{(B)}\ - \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qquad \text{(E)}\ \frac {1}{2}$

Solution

$\frac{6 @ 2}{6 \# 2} = \frac{(6)\times (2) - (2)^2}{(6) + (2) - (6) \cdot (2)^2} = \frac{8}{-16} = \frac{-1}{2} \Rightarrow \mathrm{(A)}$

2007 AMC 10A (Problems)
Preceded by Problem 1	Followed by Problem 3
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2007 AMC 10A Problems/Problem 2

From AoPSWiki

Problem

Solution

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