2003 AMC 12A Problem 10

Question

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}

Answer: D

Concepts:fraction ratio and proportion

Difficulty rating: 1440

Small hint:

The three shares are

\dfrac12,\dfrac13,\dfrac16

of the whole

Big hint:

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

Solution:

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.