2001 AMC 8 Problems/Problem 12

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 $(a\otimes b)\otimes c = \dfrac{a\otimes b + c}{a\otimes b - c}$ $(a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c}$ $(a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - ab}{a - b}}{\dfrac{a + b - ac + bc}{a - b}}$ $(a\otimes b)\otimes c = \dfrac{a + b + ac - ab}{a + b - ac + ab}$

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

$(6\otimes 4)\otimes 3 = \dfrac{6 + 4 + 18 - 12}{6 + 4 - 18 + 12}$ $(6\otimes 4)\otimes 3 = \dfrac{16}{4}$ $(6\otimes 4)\otimes 3 = 4$ $\boxed{\text {(A)}$ (Error compiling LaTeX. Unknown error_msg)

~megaboy6679

Video Solution-Cooler Method

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

2001 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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2001 AMC 8 Problems/Problem 12

Contents

Problem

Solution 1

Solution 2 (Overkill)

Video Solution-Cooler Method

See Also