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. $6#$ (Error compiling LaTeX. Unknown error_msg)2 is equal to $6+2-6*2^2$ which is $8-24 = -16$ . Therefore ${frac}(6@2)/(6#2)$ (Error compiling LaTeX. Unknown error_msg) must be equal to $8/-16 = -1/2$ . Therefore the solution is $A$ .

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2007 AMC 10A Problems/Problem 2

Problem

Solution

See also