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Difference between revisions of "2005 AMC 10A Problems/Problem 2"

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==Problem==
 
==Problem==
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For each pair of real numbers <math>a</math><math>\neq</math><math>b</math>, define the [[operation]] <math>\star</math> as
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For each pair of real numbers <math>a \neq 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>?
  
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<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>
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<math> \textbf{(A) } -\frac{2}{3}\qquad \textbf{(B) } -\frac{1}{5}\qquad \textbf{(C) } 0\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \textrm{This\, value\, is\, not\, defined.} </math>
  
 
==Solution==
 
==Solution==
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<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>
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<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 \boxed{\textbf{(C) }0}</math>
  
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==See Also==
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==Video Solution==
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*[[2005 AMC 10A Problems]]
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CHECK OUT Video Solution: https://youtu.be/5g_m3_nck8E
  
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*[[2005 AMC 10A Problems/Problem 1|Previous Problem]]
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==Video Solution 2==
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https://youtu.be/6FnnFTWUJ0s
  
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*[[2005 AMC 10A Problems/Problem 3|Next Problem]]
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~Charles3829
  
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[[Category:Introductory Algebra Problems]]
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==See also==
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{{AMC10 box|year=2005|ab=A|num-b=1|num-a=3}}
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 18:05, 25 December 2022

Problem

For each pair of real numbers $a \neq b$, define the operation $\star$ as

$(a \star b) = \frac{a+b}{a-b}$.

What is the value of $((1 \star 2) \star 3)$?

$\textbf{(A) } -\frac{2}{3}\qquad \textbf{(B) } -\frac{1}{5}\qquad \textbf{(C) } 0\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \textrm{This\, value\, is\, not\, defined.}$

Solution

$((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \boxed{\textbf{(C) }0}$

Video Solution

CHECK OUT Video Solution: https://youtu.be/5g_m3_nck8E

Video Solution 2

https://youtu.be/6FnnFTWUJ0s

~Charles3829

See also

2005 AMC 10A (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AMC 10 Problems and Solutions

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