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| − | == Problem ==
| + | Please navigate here for the solution: http://www.artofproblemsolving.com/Wiki/index.php/2001_AMC_12_Problems/Problem_3 |
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| − | The state income tax where Kristin lives is levied at the rate of <math> p\% </math> of the first <math> \</math><math>28000 </math> of annual income plus <math> (p + 2)\% </math> of any amount above <math> \</math><math>28000 </math>. Kristin noticed that the state income tax she paid amounted to <math> (p + 0.25)\% </math> of her annual income. What was her annual income?
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| − | <math> \textbf{(A) }\</math><math>28000\qquad\textbf{(B) }</math>\<math>32000\qquad\textbf{(C) }\</math><math>35000\qquad\textbf{(D) }</math>\<math>42,000\qquad\textbf{(E) }\</math><math>56000 </math>
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| − | == Solution ==
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| − | <math> 28000 \cdot \frac{p}{100}+(x-28000) \cdot \frac{p+2}{100}= \frac{p+0.25}{100}(x) </math>
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| − | <math> 280p+\frac{xp}{100}+\frac{x}{50}-280p-560 = \frac{xp}{100}+\frac{x}{400} </math>
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| − | <math> \frac{x}{50}-560 = \frac{x}{400} </math>
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| − | <math> 8x-(560)(400) = x </math>
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| − | <math> 7x = (560)(400) </math>
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| − | <math> x=(80)(400) </math>
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| − | <math> x = 32000 </math>
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| − | <math> \boxed{B} </math>.
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Revision as of 16:10, 16 March 2011
