AMC10 2001 A

AMC10 2001 A · Q9

AMC10 2001 A · Q9. It mainly tests Linear equations, Interest / growth (simple).

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

Kristin 所在州的所得税税率为年收入前 $28000$ 美元的部分按 $p\%$ 征收，超过 $28000$ 美元的部分按 $(p + 2)\%$ 征收。Kristin 注意到她缴纳的州所得税相当于她年收入的 $(p + 0.25)\%$。她的年收入是多少？

(A) $28000$ $28000$

(B) $32000$ $32000$

(D) $42000$ $42000$

(E) $56000$ $56000$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) If Kristin’s annual income is $x \ge 28{,}000$ dollars, then $\dfrac{p}{100}\cdot 28{,}000+\dfrac{p+2}{100}\cdot (x-28{,}000)=\dfrac{p+0.25}{100}\cdot x.$ Multiplying by 100 and expanding yields $28{,}000p+px+2x-28{,}000p-56{,}000=px+0.25x.$ So, $1.75x=\dfrac{7}{4}x=56{,}000$ and $x=32{,}000.$

（B）如果 Kristin 的年收入为 $x \ge 28{,}000$ 美元，则 $\dfrac{p}{100}\cdot 28{,}000+\dfrac{p+2}{100}\cdot (x-28{,}000)=\dfrac{p+0.25}{100}\cdot x.$ 两边同乘 100 并展开得到 $28{,}000p+px+2x-28{,}000p-56{,}000=px+0.25x.$ 所以，$1.75x=\dfrac{7}{4}x=56{,}000$，并且 $x=32{,}000.$

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