Difference between revisions of "2014 AIME II Problems/Problem 1"
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| − | From the given information, we can see that Abe can paint <math>\frac{1}{15}</math> of the room in an hour, Bea can paint <math>\frac{1}{15}\times\frac{3}{2} = \frac{1}{10}</math> of the room in an hour, and Coe can paint the room in <math>\frac{1}{15}\times 2 = \frac{2}{15}</math> of the room in an hour. After <math>90</math> minutes, Abe has painted <math>\frac{1}{15}\times\frac{3}{2}=\frac{1}{10}</math> of the room. Working together, Abe and Bea can paint <math>\frac{1}{15}+\frac{1}{10}=\frac{1}{6}</math> of the room, so it takes then <math>\frac{2}{5}\div \frac{1}{6}= \frac{12}{5}</math> hours to finish the first half of the room. All three working together can paint <math>\frac{1}{6}+\frac{2}{15}=\frac{3}{10}</math> of the room, and it takes them <math>\frac{1}{2}\div \frac{3}{10}=\frac{5}{3}</math> hours to finish the room. The total amount of time they take is | + | From the given information, we can see that Abe can paint <math>\frac{1}{15}</math> of the room in an hour, Bea can paint <math>\frac{1}{15}\times\frac{3}{2} = \frac{1}{10}</math> of the room in an hour, and Coe can paint the room in <math>\frac{1}{15}\times 2 = \frac{2}{15}</math> of the room in an hour. After <math>90</math> minutes, Abe has painted <math>\frac{1}{15}\times\frac{3}{2}=\frac{1}{10}</math> of the room. Working together, Abe and Bea can paint <math>\frac{1}{15}+\frac{1}{10}=\frac{1}{6}</math> of the room in an hour, so it takes then <math>\frac{2}{5}\div \frac{1}{6}= \frac{12}{5}</math> hours to finish the first half of the room. All three working together can paint <math>\frac{1}{6}+\frac{2}{15}=\frac{3}{10}</math> of the room in an hour, and it takes them <math>\frac{1}{2}\div \frac{3}{10}=\frac{5}{3}</math> hours to finish the room. The total amount of time they take is |
<cmath>90+\frac{12}{5}\times 60+\frac{5}{3}\times 60 = 90+ 144 + 100 = \boxed{334} \text{\ minutes.}</cmath> | <cmath>90+\frac{12}{5}\times 60+\frac{5}{3}\times 60 = 90+ 144 + 100 = \boxed{334} \text{\ minutes.}</cmath> | ||
Revision as of 14:40, 28 June 2014
Problem
Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room.
Solution
From the given information, we can see that Abe can paint 
of the room in an hour, Bea can paint 
of the room in an hour, and Coe can paint the room in 
of the room in an hour. After 
minutes, Abe has painted 
of the room. Working together, Abe and Bea can paint 
of the room in an hour, so it takes then 
hours to finish the first half of the room. All three working together can paint 
of the room in an hour, and it takes them 
hours to finish the room. The total amount of time they take is
![\[90+\frac{12}{5}\times 60+\frac{5}{3}\times 60 = 90+ 144 + 100 = \boxed{334} \text{\ minutes.}\]](http://latex.artofproblemsolving.com/d/4/a/d4ab50e87def3efddd7a549d665ff6c45cabcf21.png)
See also
| 2014 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
