1978 AMC 12 第 5 题

Question

四个男孩以

\$60

买了一艘船。第一个男孩支付的金额是其余三人付款总额的一半；第二个男孩支付的是其余三人付款总额的三分之一；第三个男孩支付的是其余三人付款总额的四分之一。第四个男孩支付了多少钱？

Four boys bought a boat for

\$60.

The first boy paid one half of the sum of the amounts paid by the other boys; the second boy paid one third of the sum of the amounts paid by the other boys; and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay?

\$10

\$12

\$13

\$14

\$15

答案：C

知识点：一次方程比与比例钱币

难度评级：1280

小提示：

将每处“其余男孩的付款总额”改写为

60

减去该男孩的付款额

Replace each “sum paid by the other boys” by

60

minus that boy’s payment

大提示：

分别求出前三人的付款额，再用

60

减去它们的和

Solve separately for the first three payments, then subtract their sum from

60

解答：

设前三人的付款额分别为

w

、

x

和

y

。则

\begin{aligned} w&=\frac12(60-w),\\ x&=\frac13(60-x),\\ y&=\frac14(60-y)\text{。} \end{aligned}

因此

w=20

、

x=15

、

y=12

。第四个男孩支付了

60-20-15-12=13

美元。

因此，正确答案是 C。

Let the first three payments be

w,

x,

and

y.

Then

\begin{aligned} w&=\frac12(60-w),\\ x&=\frac13(60-x),\\ y&=\frac14(60-y). \end{aligned}

Hence

w=20,

x=15,

y=12.

The fourth payment is

60-20-15-12=13

dollars.

Therefore, the correct answer is C.

1978 AMC 12 第 5 题

答案：C

难度评级：1280

小提示：

大提示：

解答：

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