AMC8 1992

AMC8 1992 · Q25

AMC8 1992 · Q25. It mainly tests Exponents & radicals, Fractions.

One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, etc. After how many pourings does exactly one tenth of the original water remain?

从一个满的容器中倒出一半的水。然后倒出剩余水的三分之一。继续这个过程：第三次倒出剩余水的四分之一，第四次倒出剩余水的五分之一，等等。经过多少次倒水后，剩余的正好是原来水的十分之一？

(A) 6 6

(B) 7 7

(D) 9 9

(E) 10 10

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): After the first pouring, $\frac{1}{2}$ remains. After the second pouring $\frac{1}{2}\times\frac{2}{3}$ remains. After the third pouring $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}$ remains. How many pouring until $\frac{1}{10}$ remains? $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}\times\frac{8}{9}\times\frac{9}{10}=\frac{1}{10}$ indicates 9 pourings.

答案（D）：第一次倒出后，剩下 $\frac{1}{2}$。第二次倒出后，剩下 $\frac{1}{2}\times\frac{2}{3}$。第三次倒出后，剩下 $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}$。要倒出多少次才会剩下 $\frac{1}{10}$？ $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}\times\frac{8}{9}\times\frac{9}{10}=\frac{1}{10}$ 这表示需要倒出 9 次。

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