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

Latest revision as of 13:10, 22 December 2024

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 1

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

Solution 2 (Overkill)

When you expand the general form of $(a\otimes b)\otimes c$ , you get $(a\otimes b)\otimes c = \dfrac{a\otimes b + c}{a\otimes b - c}$ $(a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c}$ $(a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - bc}{a - b}}{\dfrac{a + b - ac + bc}{a - b}}$ $(a\otimes b)\otimes c = \dfrac{a + b + ac -bc}{a + b - ac + bc}$

Now, substituting $a=6$ , $b=4$ , and $c=3$ :

$(6\otimes 4)\otimes 3 = \dfrac{6 + 4 + 18 - 12}{6 + 4 - 18 + 12}$ $(6\otimes 4)\otimes 3 = \dfrac{16}{4}$ $(6\otimes 4)\otimes 3 = 4$ $\boxed{\text {(A)}}$

~megaboy6679

Video Solution-Cooler Method

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

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

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

@@ Line 14: / Line 14: @@
 <cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c} </cmath>
 <cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - bc}{a - b}}{\dfrac{a + b - ac + bc}{a - b}} </cmath>
-<cmath> (a\otimes b)\otimes c = \dfrac{a + b + ac - ab}{a + b - ac + ab} </cmath>
+<cmath> (a\otimes b)\otimes c = \dfrac{a + b + ac -bc}{a + b - ac + bc} </cmath>
 Now, substituting <math>a=6</math>, <math>b=4</math>, and <math>c=3</math>:

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