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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 \neq b</math>, define the operation <math>\star</math> as


<math> (a \star b) = \frac{a+b}{a-b} </math>.
<cmath>(a \star b) = \frac{a+b}{a-b}.</cmath>


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) \ } \textrm{This\, value\, is\, not\, defined.} </math>
<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) } \text{This value is not defined.}
</math>


==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>
<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} = \boxed{\textbf{(C) } 0}</math>.


==See Also==
==Video Solution 1==
*[[2005 AMC 10A Problems]]
https://youtu.be/5g_m3_nck8E


*[[2005 AMC 10A Problems/Problem 1|Previous Problem]]
==Video Solution 2==
https://youtu.be/6FnnFTWUJ0s


*[[2005 AMC 10A Problems/Problem 3|Next Problem]]
~Charles3829


[[Category:Introductory Algebra Problems]]
==See also==
{{AMC10 box|year=2005|ab=A|num-b=1|num-a=3}}
 
[[Category:Introductory Number Theory Problems]]
{{MAA Notice}}

Latest revision as of 16:00, 1 July 2025

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 $((1 \star 2) \star 3)$?

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

Solution

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

Video Solution 1

https://youtu.be/5g_m3_nck8E

Video Solution 2

https://youtu.be/6FnnFTWUJ0s

~Charles3829

See also

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

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