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

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==Solution==
 
==Solution==
<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) }}</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 11:30, 7 December 2021

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

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

Video Solution

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

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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