Skip to main content

2003 AMC 12A Problem 10

Problem 10 of 25EasierArithmetic

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?

Answer choices

Show solution

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 12⋅23⋅56=518,\dfrac12\cdot\dfrac23\cdot\dfrac56=\dfrac{5}{18}, regardless of the order. Thus, the correct answer is D.
AoPS wiki

Tagged: fraction · ratio and proportion

More practice