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Difference between revisions of "2015 AMC 10B Problems/Problem 7"

(Created page with "==Problem== Consider the operation "minus the reciprocal of," defined by <math>a\diamond b=a-\frac{1}{b}</math>. What is <math>((1\diamond2)\diamond3)-(1\diamond(2\diamond3))<...")
 
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<math>1\diamond2=1-\dfrac{1}{2}=\dfrac{1}{2}</math>, so <math>(1\diamond2)\diamond3=\dfrac{1}{2}\diamond3=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}</math>. Also, <math>2\diamond3=2-\dfrac{1}{3}=\dfrac{5}{3}</math>, so <math>1\diamond(2\diamond3)=1-\dfrac{1}{5/3}=1-\dfrac{3}{5}=\dfrac{2}{5}</math>. Thus, <math>((1\diamond2)\diamond3)-(1\diamond(2\diamond3))=\boxed{\mathbf{(A)}\ -\dfrac{7}{30}}</math>
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<math>1\diamond2=1-\dfrac{1}{2}=\dfrac{1}{2}</math>, so <math>(1\diamond2)\diamond3=\dfrac{1}{2}\diamond3=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}</math>. Also, <math>2\diamond3=2-\dfrac{1}{3}=\dfrac{5}{3}</math>, so <math>1\diamond(2\diamond3)=1-\dfrac{1}{5/3}=1-\dfrac{3}{5}=\dfrac{2}{5}</math>. Thus, <math>((1\diamond2)\diamond3)-(1\diamond(2\diamond3))=\dfrac{1}{6}-\dfrac{2}{5}=\boxed{\mathbf{(A)}\ -\dfrac{7}{30}}</math>

Revision as of 18:13, 4 March 2015

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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