2003 AMC 12A Problem 10

Attempt Problem 10 of the 2003 AMC 12A below, then check your answer against the professionally curated solution 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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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}

Answer: D
Concepts:fractionratio and proportion
Difficulty rating: 1440
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.

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