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 (6@2)/(6#2) must be equal to 8/-16 = -1/2. Therefore the solution is E.

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 10 Problems and Solutions

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2007 AMC 10A Problems/Problem 2

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