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

## 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 $\boxed{\textbf{(B) }\textdollar 32000}$.

## Solution 2

Let $A$, $T$ be Kristin's annual income and the income tax total, respectively. Notice that \begin{align*} T &= p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) \\ &= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000) \\ &= p\%\cdot A + 2\%\cdot(A - 28000) \end{align*} We are also given that $$T = (p + 0.25)\%\cdot A = p\%\cdot A + 0.25\%\cdot A$$ Thus, $$p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cdot A + 0.25\%\cdot A$$ $$2\%\cdot(A - 28000) = 0.25\%\cdot A$$ Solve for $A$ to obtain $A = \boxed{\textbf{(B) }\textdollar 32000}$.

~ Nafer

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