Difference between revisions of "2000 AMC 8 Problems/Problem 17"

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==Problem==
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The operation <math> \otimes </math> is defined for all nonzero numbers by <math> a\otimes b =\frac{a^{2}}{b} </math>. Determine <math> [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] </math>.
 
The operation <math> \otimes </math> is defined for all nonzero numbers by <math> a\otimes b =\frac{a^{2}}{b} </math>. Determine <math> [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] </math>.
  
 
<math> \text{(A)}\ -\frac{2}{3}\qquad\text{(B)}\ -\frac{1}{4}\qquad\text{(C)}\ 0\qquad\text{(D)}\ \frac{1}{4}\qquad\text{(E)}\ \frac{2}{3} </math>
 
<math> \text{(A)}\ -\frac{2}{3}\qquad\text{(B)}\ -\frac{1}{4}\qquad\text{(C)}\ 0\qquad\text{(D)}\ \frac{1}{4}\qquad\text{(E)}\ \frac{2}{3} </math>
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==Solution==
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Follow PE(MD)(AS), doing the innermost parentheses first.
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<math> [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] </math>
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<math> [\frac{1^2}{2}\otimes 3]-[1\otimes \frac{2^2}{3}] </math>
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<math> [\frac{1}{2}\otimes 3]-[1\otimes \frac{4}{3}] </math>
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<math> [\frac{(\frac{1}{2})^{2}}{3}]-[\frac{1^2}{(\frac{4}{3})}] </math>
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<math> [\frac{1}{4} \cdot \frac{1}{3}]-[\frac{3}{4}] </math>
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<math>\frac{1}{12} - \frac{3}{4}</math>
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<math>\frac{1}{12} - \frac{9}{12}</math>
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<math>\frac{-8}{12}</math>
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<math>-\frac{2}{3}</math>, which is answer <math>\boxed{A}</math>
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==See Also==
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{{AMC8 box|year=2000|num-b=16|num-a=18}}

Revision as of 20:45, 30 July 2011

Problem

The operation $\otimes$ is defined for all nonzero numbers by $a\otimes b =\frac{a^{2}}{b}$. Determine $[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)]$.

$\text{(A)}\ -\frac{2}{3}\qquad\text{(B)}\ -\frac{1}{4}\qquad\text{(C)}\ 0\qquad\text{(D)}\ \frac{1}{4}\qquad\text{(E)}\ \frac{2}{3}$

Solution

Follow PE(MD)(AS), doing the innermost parentheses first.

$[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)]$

$[\frac{1^2}{2}\otimes 3]-[1\otimes \frac{2^2}{3}]$

$[\frac{1}{2}\otimes 3]-[1\otimes \frac{4}{3}]$

$[\frac{(\frac{1}{2})^{2}}{3}]-[\frac{1^2}{(\frac{4}{3})}]$

$[\frac{1}{4} \cdot \frac{1}{3}]-[\frac{3}{4}]$

$\frac{1}{12} - \frac{3}{4}$

$\frac{1}{12} - \frac{9}{12}$

$\frac{-8}{12}$

$-\frac{2}{3}$, which is answer $\boxed{A}$

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions