Question

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.

Answer

Solution:

Abe paints $\frac{1}{900}$ of the room per minute, so Bea paints $\frac{3}{2} \cdot \frac{1}{900} = \frac{1}{600}$ per minute and Coe paints $\frac{2}{900} = \frac{1}{450}$ per minute. In the first $90$ minutes Abe paints $\frac{90}{900} = \frac{1}{10}$ of the room.

Abe and Bea together paint $\frac{1}{900} + \frac{1}{600} = \frac{1}{360}$ per minute, and they must bring the total from $\frac{1}{10}$ up to $\frac{1}{2},$ which takes $\frac{2}{5} \cdot 360 = 144$ minutes. All three together paint $\frac{1}{360} + \frac{1}{450} = \frac{1}{200}$ per minute, so the remaining half of the room takes $\frac{1}{2} \cdot 200 = 100$ minutes.

The total time is $90 + 144 + 100 = 334$ minutes.

2014 AIME II Problem 1

Solution:

Problem 1 in Other Years