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AIME 2014 II · 第 1 题

AIME 2014 II — Problem 1

专题
Contest Math
难度
L4
来源
AIME

题目详情

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.

解析

Solution

From the given information, we can see that Abe can paint 115\frac{1}{15} of the room in an hour, Bea can paint 115×32=110\frac{1}{15}\times\frac{3}{2} = \frac{1}{10} of the room in an hour, and Coe can paint the room in 115×2=215\frac{1}{15}\times 2 = \frac{2}{15} of the room in an hour. After 9090 minutes, Abe has painted 115×32=110\frac{1}{15}\times\frac{3}{2}=\frac{1}{10} of the room. Working together, Abe and Bea can paint 115+110=16\frac{1}{15}+\frac{1}{10}=\frac{1}{6} of the room in an hour, so it takes then 25÷16=125\frac{2}{5}\div \frac{1}{6}= \frac{12}{5} hours to finish the first half of the room. All three working together can paint 16+215=310\frac{1}{6}+\frac{2}{15}=\frac{3}{10} of the room in an hour, and it takes them 12÷310=53\frac{1}{2}\div \frac{3}{10}=\frac{5}{3} hours to finish the room. The total amount of time they take is

90+125×60+53×60=90+144+100=334 minutes.90+\frac{12}{5}\times 60+\frac{5}{3}\times 60 = 90+ 144 + 100 = \boxed{334} \text{\ minutes.}

Video Solution by OmegaLearn

https://youtu.be/00Ngozqw2d0?t=214

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