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Difference between revisions of "2001 AMC 12 Problems/Problem 3"

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== Problem ==
 
== Problem ==
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The state income tax where Kristin lives is levied at the rate of <math>p%</math> of the first
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The state income tax where Kristin lives is levied at the rate of <math>p\%</math> of the first
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<dollar/><math>28000</math> of annual income plus <math>(p + 2)%</math> of any amount above <dollar/><math>28000</math>. Kristin
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<math>\textdollar 28000</math> of annual income plus <math>(p + 2)\%</math> of any amount above <math>\textdollar 28000</math>. Kristin
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noticed that the state income tax she paid amounted to <math>(p + 0.25)%</math> of her
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noticed that the state income tax she paid amounted to <math>(p + 0.25)\%</math> of her
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annual income. What was her annual income?
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annual income. What was her annual income?  
  
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<math>\text{(A)}\,</math><dollar/><math>28000\qquad \text{(B)}\,</math><dollar/><math>32000\qquad \text{(C)}\,</math><dollar/><math>35000\qquad \text{(D)}\,</math><dollar/><math>42000\qquad \text{(E)}\,</math><dollar/><math>56000</math>
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<math>\text{(A)}\,\textdollar 28000 \qquad \text{(B)}\,\textdollar 32000 \qquad \text{(C)}\,\textdollar 35000 \qquad \text{(D)}\,\textdollar 42000 \qquad \text{(E)}\,\textdollar 56000</math>
  
 
== Solution==
 
== Solution==

Revision as of 19:35, 13 June 2015

The following problem is from both the 2001 AMC 12 #3 and 2001 AMC 10 #9, so both problems redirect to this page.

Problem

The state income tax where Kristin lives is levied at the rate of $p\%$ of the first $\textdollar 28000$ of annual income plus $(p + 2)\%$ of any amount above $\textdollar 28000$. Kristin noticed that the state income tax she paid amounted to $(p + 0.25)\%$ of her annual income. What was her annual income?

$\text{(A)}\,\textdollar 28000 \qquad \text{(B)}\,\textdollar 32000 \qquad \text{(C)}\,\textdollar 35000 \qquad \text{(D)}\,\textdollar 42000 \qquad \text{(E)}\,\textdollar 56000$

Solution

Let the income amount be denoted by $A$.

We know that $\frac{A(p+.25)}{100}=\frac{28000p}{100}+\frac{(p+2)(A-28000)}{100}$.

We can now try to solve for $A$:

$(p+.25)A=28000p+Ap+2A-28000p-56000$

$.25A=2A-56000$

$A=32000$

So the answer is $\boxed{B}$

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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