Difference between revisions of "2005 AMC 10A Problems/Problem 2"
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==Problem== | ==Problem== | ||
| − | For each pair of real numbers <math>a \neq b</math>, define the | + | For each pair of real numbers <math>a \neq b</math>, define the operation <math>\star</math> as |
| − | < | + | <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> \textbf{(A) } -\frac{2}{3}\qquad \textbf{(B) } -\frac{1}{5}\qquad \textbf{(C) } 0\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \ | + | <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} = | + | <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== | + | ==Video Solution 1== |
| − | + | https://youtu.be/5g_m3_nck8E | |
==Video Solution 2== | ==Video Solution 2== | ||
Latest revision as of 17:00, 1 July 2025
Problem
For each pair of real numbers 
, define the operation 
as
![\[(a \star b) = \frac{a+b}{a-b}.\]](http://latex.artofproblemsolving.com/7/4/d/74d9344dab3ec93518b1f12276a642d901688dbb.png)
What is the value of 
?
Solution
.
Video Solution 1
Video Solution 2
~Charles3829
See also
| 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. 
