2005 AMC 10A Problems/Problem 2: Difference between revisions

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

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 6: / Line 6: @@
 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) } \ \text{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==
-<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{\mathrm{(C) } \ 0}</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>.
 ==Video Solution 1==

Page

Toolbox

Search