AMC12 2003 A

AMC12 2003 A · Q10

AMC12 2003 A · Q10. It mainly tests Fractions, Logic puzzles.

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 the correct share of candy, what fraction of the candy goes unclaimed?

Al、Bert 和 Carl 是学校一次万圣节糖果堆抽奖的获胜者，他们将分别按 $3:2:1$ 的比例分糖果。由于一些混乱，他们在不同时间来领取奖品，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正确的那份糖果，那么有多少比例的糖果无人领取？

(A) \frac{1}{18} \frac{1}{18}

(B) \frac{1}{6} \frac{1}{6}

(D) \frac{5}{18} \frac{5}{18}

(E) \frac{5}{12} \frac{5}{12}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Because the ratios are $3:2:1$, Al, Bert, and Carl believe that they need to take $1/2$, $1/3$, and $1/6$ of the pile when they each arrive, respectively. After each person comes, $1/2$, $2/3$, and $5/6$ of the pile's size (just before each came) remains. The pile starts at $1$, and at the end $\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6}\cdot 1 = \frac{5}{18}$ of the original pile goes unclaimed. (Note that because of the properties of multiplication, it does not matter what order the three come in.) Hence the answer is $\boxed{\mathrm{(D)}\ \dfrac{5}{18}}$.

因为比例是 $3:2:1$，所以 Al、Bert 和 Carl 分别认为自己需要拿走糖果堆的 $1/2$、$1/3$ 和 $1/6$。每个人来过之后，糖果堆（相对于他来之前的大小）分别剩下 $1/2$、$2/3$ 和 $5/6$。糖果堆初始为 $1$，最后无人领取的部分为 $\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6}\cdot 1 = \frac{5}{18}$。（注意由于乘法的性质，三人到达的顺序并不影响结果。）因此答案是 $\boxed{\mathrm{(D)}\ \dfrac{5}{18}}$。

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