Difference between revisions of "2005 AMC 10B Problems/Problem 3"

 
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== Problem ==
 
== Problem ==
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A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?
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<math>\mathrm{(A)} \frac{1}{10} \qquad \mathrm{(B)} \frac{1}{9} \qquad \mathrm{(C)} \frac{1}{3} \qquad \mathrm{(D)} \frac{4}{9} \qquad \mathrm{(E)} \frac{5}{9} </math>
 
== Solution ==
 
== Solution ==
== See Also ==
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After the first day, there is <math>1-\left(\dfrac{1}{3}\times1\right)</math> gallons left, or <math>\dfrac{2}{3}</math> gallons. After the second day, there is a total of <math>\dfrac{2}{3}-\left(\dfrac{1}{3}\times\dfrac{2}{3}\right)=\dfrac{2}{3}-\dfrac{2}{9}=\dfrac{4}{9}</math>. Therefore, the fraction of the original amount of paint that is left is <math>\dfrac{\dfrac{4}{9}}{1}=\boxed{\mathrm{(D)}\,\dfrac{4}{9}}</math>
*[[2005 AMC 10B Problems]]
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==See Also ==

Revision as of 15:35, 28 June 2011

Problem

A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?

$\mathrm{(A)} \frac{1}{10} \qquad \mathrm{(B)} \frac{1}{9} \qquad \mathrm{(C)} \frac{1}{3} \qquad \mathrm{(D)} \frac{4}{9} \qquad \mathrm{(E)} \frac{5}{9}$

Solution

After the first day, there is $1-\left(\dfrac{1}{3}\times1\right)$ gallons left, or $\dfrac{2}{3}$ gallons. After the second day, there is a total of $\dfrac{2}{3}-\left(\dfrac{1}{3}\times\dfrac{2}{3}\right)=\dfrac{2}{3}-\dfrac{2}{9}=\dfrac{4}{9}$. Therefore, the fraction of the original amount of paint that is left is $\dfrac{\dfrac{4}{9}}{1}=\boxed{\mathrm{(D)}\,\dfrac{4}{9}}$

See Also

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