2007 AMC 10A Problems/Problem 2

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

$6@2$ must be equal to $6*2-2^2$ which is 8. $6\# 2$ is equal to $6+2-6*2^2$ which is $8-24 = -16$. Therefore $\frac{6@2}{6\# 2}$ must be equal to $\frac{8}{-16} = -\frac{1}{2}$. Therefore the solution is $A$.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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