2003 AMC 12A 第 10 题

Question

Al、Bert 和 Carl 在学校抽奖中赢得一堆万圣节糖果，按

3 : 2 : 1

的比例分给他们。由于混乱，他们在不同时间来领奖，并且每个人都以为自己是第一个到的。如果每个人都拿走他认为自己应得的份额，那么有多少比例的糖果无人领取？

Al, Bert, and Carl are the winners of a school drawing for a pile of Halloween candy, which they are to divide in a ratio of

3 : 2 : 1,

respectively. Due to some confusion they come at different times to claim their prizes, and each assumes he is the first to arrive. If each takes what he believes to be his correct share of candy, what fraction of the candy goes unclaimed?

\dfrac{1}{18}

\dfrac{1}{6}

\dfrac{2}{9}

\dfrac{5}{18}

\dfrac{5}{12}

答案：D

知识点：分数比与比例

难度评级：1440

小提示：

三人的份额分别是整堆的

\dfrac12,\dfrac13,\dfrac16

。

The three shares are

\dfrac12,\dfrac13,\dfrac16

of the whole

大提示：

每个人都会留下当时糖果的固定比例，所以顺序不影响结果。

Each person leaves a fixed fraction of whatever candy is present, so the order does not matter

解答：

三人的份额分别是

\dfrac12,\dfrac13,\dfrac16

。

每个人都以为自己是第一个到的，所以在各自到达时，Al 留下当时糖果的

\dfrac12

，Bert 留下

\dfrac23

，Carl 留下

\dfrac56

。

无论三人按什么顺序到达，无人领取的比例都是

\dfrac12\cdot\dfrac23\cdot\dfrac56=\dfrac{5}{18}

。

所以正确答案是 D。

The shares are

\dfrac12,\dfrac13,\dfrac16

of the pile.

Each person assumes he is first, so Al leaves

\dfrac12,

Bert leaves

\dfrac23,

and Carl leaves

\dfrac56

of the candy present when he arrives.

The unclaimed fraction is

\dfrac12\cdot\dfrac23\cdot\dfrac56=\dfrac{5}{18},

regardless of the order.

Thus, the correct answer is D.

2003 AMC 12A 第 10 题

答案：D

难度评级：1440

小提示：

大提示：

解答：

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