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>.
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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 ==
*[[2003 AMC 12A Problems]]
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{{AMC12 box|year=2003|ab=A|num-b=9|num-a=11}}
*[[2003 AMC 12A/Problem 9|Previous Problem]]
 
*[[2003 AMC 12A/Problem 11|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 10:16, 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 $3:2:1$, 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?

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

Solution

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 $\boxed{\mathrm{(D)}\ \dfrac{5}{18}}$.

See Also

2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 12 Problems and Solutions

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