Question

从下列集合中选出四个不同数字 $W, X, Y$ 和 $Z$ ：

$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}.$

如果要使 $\dfrac{W}{X} + \dfrac{Y}{Z}$ 尽可能小，那么 $\dfrac{W}{X} + \dfrac{Y}{Z}$ 必须等于

Let $W, X, Y,$ and $Z$ be four different digits selected from the set

$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}.$

If the sum $\dfrac{W}{X} + \dfrac{Y}{Z}$ is to be as small as possible, then $\dfrac{W}{X} + \dfrac{Y}{Z}$ must equal

$\dfrac{2}{17}$

$\dfrac{3}{17}$

$\dfrac{17}{72}$

$\dfrac{25}{72}$

$\dfrac{13}{36}$

答案：D

知识点：分数最优化

难度评级：1000

解答：

小分子和大分母会产生较小分数，所以用 $1$ 和 $2$ 作分子，用 $8$ 和 $9$ 作分母。

把较大的分子配给较大的分母，得到 $\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72}$ ，它小于 $\dfrac19 + \dfrac28 = \dfrac{26}{72}$ 。所以最小和是 $\dfrac{25}{72}$ 。

所以正确答案是 D。

Small numerators and large denominators make small fractions, so use $1$ and $2$ as numerators and $8$ and $9$ as denominators.

Pairing the larger numerator with the larger denominator gives $\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72},$ which is smaller than $\dfrac19 + \dfrac28 = \dfrac{26}{72}.$ So the minimum sum is $\dfrac{25}{72}.$

Thus, the correct answer is D .

1994 AMC 8 第 20 题

答案：D

难度评级：1000

解答：

其他年份的第 20 题