Difference between revisions of "2001 AMC 8 Problems/Problem 12"

(Gave Solution)
 
(Solution)
(2 intermediate revisions by 2 users not shown)
Line 9: Line 9:
 
<math> 6\otimes4=\frac{6+4}{6-4}=5 </math>.  
 
<math> 6\otimes4=\frac{6+4}{6-4}=5 </math>.  
 
<math> 5\otimes3=\frac{5+3}{5-3}=4, \boxed{\text{A}} </math>
 
<math> 5\otimes3=\frac{5+3}{5-3}=4, \boxed{\text{A}} </math>
 +
 +
==Video Solution-Cooler Method==
 +
https://www.youtube.com/watch?v=ZfwtAiH_6PI&t=36s
  
 
==See Also==
 
==See Also==
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 
{{AMC8 box|year=2001|num-b=11|num-a=13}}
 
{{AMC8 box|year=2001|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Revision as of 20:42, 26 December 2023

Problem

If $a\otimes b = \dfrac{a + b}{a - b}$, then $(6\otimes 4)\otimes 3 =$

$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$

Solution

$6\otimes4=\frac{6+4}{6-4}=5$. $5\otimes3=\frac{5+3}{5-3}=4, \boxed{\text{A}}$

Video Solution-Cooler Method

https://www.youtube.com/watch?v=ZfwtAiH_6PI&t=36s

See Also

2001 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 AJHSME/AMC 8 Problems and Solutions

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