2005 AMC 10A Problems/Problem 2: Difference between revisions
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==Problem== | ==Problem== | ||
For each pair of real numbers <math>a</math><math>\neq</math><math>b</math>, define the operation <math>\star</math> as | For each pair of real numbers <math>a</math><math>\neq</math><math>b</math>, define the [[operation]] <math>\star</math> as | ||
<math> (a \star b) = \frac{a+b}{a-b} </math>. | <math> (a \star b) = \frac{a+b}{a-b} </math>. | ||
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What is the value of <math> ((1 \star 2) \star 3)</math>? | What is the value of <math> ((1 \star 2) \star 3)</math>? | ||
<math> \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } This value is not defined. </math> | <math> \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} </math> | ||
==Solution== | ==Solution== | ||
<math> ((1 \star 2) \star 3) = (\frac{1+2}{1-2} \star 3) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \ | <math> ((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \mathrm{(C)}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 09:34, 2 August 2006
Problem
For each pair of real numbers 
, define the operation 
as
.
What is the value of 
?
Solution
