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

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

### Solution 1

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 $\boxed{B}$

### Solution 2

Let $A$, $T$ be Kristin's annual income. Notice that

\begin{align*}
p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) (Error compiling LaTeX. ! LaTeX Error: \begin{align*} on input line 20 ended by \end{document}.)

$$= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000)$$ $$= p\%\cdot A + 2\%\cdot(A - 28000)$$

\begin{align*} ((2x+3)^3)' &= 3(2x+3)^2 \cdot (2x+3)' \\ &= 3(2x+3)^2 \cdot 2 \\ &= 6(2x+3)^2. \end{align*}

## See Also

 2001 AMC 12 (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 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 byProblem 8 Followed byProblem 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.

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