Difference between revisions of "2003 AMC 12A Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | Because the ratios are <math>3:2:1</math>, Al, Bert, and Carl believe that they need to take <math>1/2</math>, <math>1/3</math>, and <math>1/6</math> of the pile when they each arrive, respectively. After each person comes, <math>1/2</math>, <math>2/3</math>, and <math>5/6</math> of the pile's size (just before each came) remains. The pile starts at <math>1</math>, and at the end <math>\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6}\cdot 1 = \frac{5}{18}</math> 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 <math>D</math>. | + | Because the ratios are <math>3:2:1</math>, Al, Bert, and Carl believe that they need to take <math>1/2</math>, <math>1/3</math>, and <math>1/6</math> of the pile when they each arrive, respectively. After each person comes, <math>1/2</math>, <math>2/3</math>, and <math>5/6</math> of the pile's size (just before each came) remains. The pile starts at <math>1</math>, and at the end <math>\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6}\cdot 1 = \frac{5}{18}</math> 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 <math>\boxed{\mathrm{(D)}\ \dfrac{5}{18}}</math>. |
== See Also == | == See Also == | ||
| − | + | {{AMC12 box|year=2003|ab=A|num-b=9|num-a=11}} | |
| − | |||
| − | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:16, 4 July 2013
Problem
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 
, 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?
Solution
Because the ratios are , Al, Bert, and Carl believe that they need to take 
, 
, and 
of the pile when they each arrive, respectively. After each person comes, , 
, and 
of the pile's size (just before each came) remains. The pile starts at 
, and at the end 
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 
.
See Also
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
