Difference between revisions of "2004 AMC 10A Problems/Problem 2"

Revision as of 17:27, 20 July 2014

Problem

For any three real numbers $a$ , $b$ , and $c$ , with $b\neq c$ , the operation $\otimes$ is defined by: $\otimes(a,b,c)=\frac{a}{b-c}$ What is $\otimes(\otimes(1,2,3),\otimes(2,3,1),\otimes(3,1,2))$ ?

$\mathrm{(A) \ } -\frac{1}{2}\qquad \mathrm{(B) \ } -\frac{1}{4} \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac{1}{4} \qquad \mathrm{(E) \ } \frac{1}{2}$

Solution

$\otimes(\frac{1}{2-3},\frac{2}{3-1},\frac{3}{1-2})=\otimes(-1,1,-3)=\frac{-1}{1+3}=-\frac{1}{4}\Longrightarrow\boxed{\mathrm{(B)}\ -\frac{1}{4}}$

2004 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 == Solution ==
-<math>\otimes<cmath> \left(\frac{1}{2-3}, \frac{2}{3-1}, \frac{3}{1-2}\right)=</cmath>\otimes</math><math>(-1,1,-3)=\frac{-1}{1+3}=-\frac{1}{4}\Longrightarrow\mathrm{(B)}</math>
+<math>\otimes(\frac{1}{2-3},\frac{2}{3-1},\frac{3}{1-2})=\otimes(-1,1,-3)=\frac{-1}{1+3}=-\frac{1}{4}\Longrightarrow\boxed{\mathrm{(B)}\ -\frac{1}{4}}</math>
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Difference between revisions of "2004 AMC 10A Problems/Problem 2"

Revision as of 17:27, 20 July 2014

Problem

Solution

See also