2001 AMC 10 Problem 9

Let Kristin's income be $I$ dollars. If $I \le 28000$ her rate would be exactly $p\%$, not $(p+0.25)\%$, so $I > 28000$. Her tax, in dollars, is
$$\frac{p}{100}\cdot 28000 + \frac{p+2}{100}(I - 28000).$$
Regroup: this equals $\frac{p}{100}\cdot I + \frac{2}{100}(I - 28000)$, i.e. $p\%$ of everything plus an extra $2\%$ of the part above $28000$.

The tax is also $\frac{p+0.25}{100} I = \frac{p}{100} I + \frac{0.25}{100} I$. Comparing, the $p\%$ pieces cancel and
$$2(I - 28000) = 0.25 I \quad\Longrightarrow\quad 1.75 I = 56000 \quad\Longrightarrow\quad I = 32000.$$

The answer is $\boxed{\textbf{(B)}\ \$32000}$.

A bracketed tax is "base rate on everything plus a surcharge on the excess." Writing it that way makes the unknown base rate $p$ vanish, leaving a one-variable percent equation: the extra $2\%$ on the excess must equal an extra $0.25\%$ on the whole. Whenever a parameter appears on both sides, look for the cancellation before computing.

Test (B): the excess is $4000$, so the surcharge is $2\%$ of $4000 = 80$ dollars, and $0.25\%$ of $32000$ is also $80$. Other choices fail (for $42000$: $2\%$ of $14000 = 280$ versus $0.25\%$ of $42000 = 105$).

• Dropping the percent signs inconsistently (mixing 2 with 0.25%) or assuming income is at most 28000, which makes the condition impossible.
• Applying $(p+2)\%$ to the entire income rather than only to the amount above $28000$.