1970 AMC 12 第 35 题

Question

一名退休员工领取的年度养老金与其服务年数的平方根成正比。若他多服务

a

年，养老金将增加

p

美元；若他多服务

b

年（

b\ne a

），养老金将比原年度养老金增加

q

美元。用

a

、

b

、

p

、

q

表示他的年度养老金。

A retiring employee receives an annual pension proportional to the square root of the number of years of his service. Had he served

a

years more, his pension would have been

p

dollars greater, whereas, had he served

b

years more (

b\ne a

), his pension would have been

q

dollars greater than the original annual pension. Find his annual pension in terms of

a,

b,

p,

and

q.

\dfrac{p^2-q^2}{2(a-b)}

\dfrac{(p-q)^2}{2\sqrt{ab}}

\dfrac{ap^2-bq^2}{2(ap-bq)}

\dfrac{aq^2-bp^2}{2(bp-aq)}

\sqrt{(a-b)(p-q)}

答案：D

知识点：比与比例根式方程组代数变形

难度评级：2440

小提示：

设当前养老金为

X=k\sqrt n

，并写出两种假设情况下的养老金

Let the current pension be

X=k\sqrt n

and write the two hypothetical pensions

大提示：

将

X+p=k\sqrt{n+a}

和

X+q=k\sqrt{n+b}

分别平方，再消去

n

和

k^2

Square

X+p=k\sqrt{n+a}

and

X+q=k\sqrt{n+b}

, then eliminate both

n

and

k^2

解答：

设当前养老金为

X=k\sqrt n

。因为

X+p=k\sqrt{n+a}

，且

X+q=k\sqrt{n+b}

，分别平方并利用

X^2=k^2n

，得到下面前两个方程。再将它们分别乘以

b

和

a

，相减后得到第三个方程：

\begin{gathered} 2pX+p^2=k^2a,\\ 2qX+q^2=k^2b,\\ 2X(bp-aq)=aq^2-bp^2 \end{gathered}\text{。}

因此

X=\dfrac{aq^2-bp^2}{2(bp-aq)}

。

因此，正确答案是 D。

Let the current pension be

X=k\sqrt n.

Since

X+p=k\sqrt{n+a}

and

X+q=k\sqrt{n+b},

squaring and using

X^2=k^2n

gives the first line below. Multiplying its equations by

b

and

a,

respectively, and subtracting gives the second:

\begin{gathered} 2pX+p^2=k^2a,\\ 2qX+q^2=k^2b,\\ 2X(bp-aq)=aq^2-bp^2. \end{gathered}

Hence

X=\dfrac{aq^2-bp^2}{2(bp-aq)}.

Therefore, the correct answer is D.

1970 AMC 12 第 35 题

答案：D

难度评级：2440

小提示：

大提示：

解答：

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