2003 AMC 12A Problem 10

Below is the professionally curated solution for Problem 10 of the 2003 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12A solutions, or check the answer key.

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Concepts:fractionratio and proportion

Difficulty rating: 1440

10.

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 3:2:1,3 : 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?

118\dfrac{1}{18}

16\dfrac{1}{6}

29\dfrac{2}{9}

518\dfrac{5}{18}

512\dfrac{5}{12}

Solution:

The shares are 12,13,16\dfrac12,\dfrac13,\dfrac16 of the pile.

Each person assumes he is first, so Al leaves 12,\dfrac12, Bert leaves 23,\dfrac23, and Carl leaves 56\dfrac56 of the candy present when he arrives.

The unclaimed fraction is 122356=518,\dfrac12\cdot\dfrac23\cdot\dfrac56=\dfrac{5}{18}, regardless of the order.

Thus, the correct answer is D.

Problem 10 in Other Years