AMC8 1994

AMC8 1994 · Q20

AMC8 1994 · Q20. It mainly tests Fractions, Casework.

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 $\frac{W}{X} + \frac{Y}{Z}$ is to be as small as possible, then $\frac{W}{X} + \frac{Y}{Z}$ must equal

让W、X、Y和Z是从集合{1, 2, 3, 4, 5, 6, 7, 8, 9}中选的四个不同数字。如果要使 $\frac{W}{X} + \frac{Y}{Z}$ 尽可能小，那么 $\frac{W}{X} + \frac{Y}{Z}$ 必须等于

(A) $\frac{2}{17}$ $\frac{2}{17}$

(B) $\frac{3}{17}$ $\frac{3}{17}$

(D) $\frac{25}{72}$ $\frac{25}{72}$

(E) $\frac{13}{36}$ $\frac{13}{36}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

$\frac{1}{8} + \frac{2}{9} = \frac{25}{72}$, smaller than $\frac{1}{9} + \frac{2}{8} = \frac{26}{72}$.

$\frac{1}{8} + \frac{2}{9} = \frac{25}{72}$，小于 $\frac{1}{9} + \frac{2}{8} = \frac{26}{72}$。

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