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

Latest revision as of 17:37, 27 October 2024

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

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

$6\otimes4=\frac{6+4}{6-4}=\frac{10}{2}=5$ . $5\otimes3=\frac{5+3}{5-3}=\frac{8}{2}=\boxed{\textbf{(A)}\ 4}$

2001 AMC 8 (Problems • Answer Key • Resources)
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

@@ Line 7: / Line 7: @@
 ==Solution==
-<math> 6\otimes4=\frac{6+4}{6-4}=5 </math>.
+<math> 6\otimes4=\frac{6+4}{6-4}=\frac{10}{2}=5 </math>.
-<math> 5\otimes3=\frac{5+3}{5-3}=4, \boxed{\text{A}} </math>
+<math> 5\otimes3=\frac{5+3}{5-3}=\frac{8}{2}=\boxed{\textbf{(A)}\ 4} </math>
+==Video Solution-Cooler Method==
+https://www.youtube.com/watch?v=ZfwtAiH_6PI&t=36s
 ==See Also==
 {{AMC8 box|year=2001|num-b=11|num-a=13}}
+{{MAA Notice}}