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

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

2005 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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

@@ Line 13: / Line 13: @@
 ==Video Solution==
 CHECK OUT Video Solution: https://youtu.be/5g_m3_nck8E
+==Video Solution 2==
+https://youtu.be/6FnnFTWUJ0s
+~Charles3829
 ==See also==

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