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

Line 7: Line 7:
 
==See Also ==
 
==See Also ==
 
{{AMC10 box|year=2005|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2005|ab=B|num-b=2|num-a=4}}
 +
{{MAA Notice}}

Revision as of 11:14, 4 July 2013

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

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS