2003 AMC 12A Problem 10

Problem 10 of 25EasierAlgebra

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?

Show solution

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.

Answer choices

Solution

More practice