Difference between revisions of "2003 AMC 12A Problems/Problem 10"

m (2003 AMC 12A/Problem 10 moved to 2003 AMC 12A Problems/Problem 10: Was inconsistent with other page titles.)
Line 10: Line 10:
*[[2003 AMC 12A Problems]]
*[[2003 AMC 12A Problems]]
*[[2003 AMC 12A Problems/Problem 9|Previous Problem]]
*[[2003 AMC 12A Problems/Problem 9|Previous Problem]]
*[[2003 AMC 12A/Problem 11|Next Problem]]
*[[2003 AMC 12A Problems/Problem 11|Next Problem]]
[[Category:Introductory Algebra Problems]]
[[Category:Introductory Algebra Problems]]

Revision as of 12:04, 16 November 2008


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 the correct share of candy, what fraction of the candy goes unclaimed?

$\mathrm{(A) \ } \frac{1}{18}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{2}{9}\qquad \mathrm{(D) \ } \frac{5}{18}\qquad \mathrm{(E) \ } \frac{5}{12}$


Because the ratios are $3:2:1$, Al, Bert, and Carl believe that they need to take $1/2$, $1/3$, and $1/6$ of the pile when they each arrive, respectively. After each person comes, $1/2$, $2/3$, and $5/6$ of the pile's size (just before each came) remains. The pile starts at $1$, and at the end $\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6}\cdot 1 = \frac{5}{18}$ of the original pile goes unclaimed. (Note that because of the properties of multiplication, it does not matter what order the three come in.) Hence the answer is $D$.

See Also

Invalid username
Login to AoPS