2015 AMC 10B Problems/Problem 7

Problem

Consider the operation "minus the reciprocal of," defined by $a\diamond b=a-\frac{1}{b}$ . What is $((1\diamond2)\diamond3)-(1\diamond(2\diamond3))$ ?

$\textbf{(A) } -\dfrac{7}{30} \qquad\textbf{(B) } -\dfrac{1}{6} \qquad\textbf{(C) } 0 \qquad\textbf{(D) } \dfrac{1}{6} \qquad\textbf{(E) } \dfrac{7}{30}$

Solution

$1\diamond2=1-\dfrac{1}{2}=\dfrac{1}{2}$ , so $(1\diamond2)\diamond3=\dfrac{1}{2}\diamond3=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$ . Also, $2\diamond3=2-\dfrac{1}{3}=\dfrac{5}{3}$ , so $1\diamond(2\diamond3)=1-\dfrac{1}{5/3}=1-\dfrac{3}{5}=\dfrac{2}{5}$ . Thus, $((1\diamond2)\diamond3)-(1\diamond(2\diamond3))=\dfrac{1}{6}-\dfrac{2}{5}=\boxed{\mathbf{(A)}\ -\dfrac{7}{30}}$

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2015 AMC 10B Problems/Problem 7

Problem

Solution