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Project IR Butler 2019-20 - Get one IR Question EverydayQuestion # 195, Date : 14-Apr-2020
This post is a part of Project IR Butler 2019-20.
Click here for Details Perry, Maria, and Lorna are painting rooms in a college dormitory. Working alone, Perry can paint a standard room in 3 hours, Maria can paint a standard room in 2 hours, and Lorna can paint a standard room in 2 hours 30 minutes. Perry, Maria, and Lorna have decided that, to speed up the work, 2 of them will paint a standard room together.`
Select the value closest to the shortest time in which a 2-person team could paint a standard room, and select the value closest to the longest time in which a 2-person team could paint a standard room, with each person working at his or her respective rate. Make only two selections, one in each column.
Shortest time | Longest time | |
| | 49 minutes |
| | 1 hour 7 minutes |
| | 1 hour 12 minutes |
| | 1 hour 14 minutes |
| | 1 hour 22 minutes |
| | 1 hour 45 minutes |
sthahviP, M and L can do the work individually in 3, 2 and 2.5 hrs.
When two of them coworkers together:
1) Least time : It will be when the two fastest work together, that is M and L.
You can work out with 1 hr work of each, which is 1/2 and 1/2.5 or 2/5
\(\frac{1}{2}+\frac{2}{5}=9/10\). So they will finish the work in 10/9 hrs or 60*10/9 =200/3=66.67 minutes or 1 hr 7 mins approx.
OR
Take 10 hrs ( LCM of 2 and 2.5): In 10 hrs M can do 5 jobs at rate of 1 per 2 hr and L can do 4 at rate of 2.5 per hour.
Thus combined both do 4+5 or 9 in 10 hrs or 10/9 for 1 work.
(2) Maximum time : It will be when the two slowest work together, that is P and L.
You can work out with 1 hr work of each, which is 1/3 and 1/2.5 or 2/5
\(\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\). So they will finish the work in 15/11 hrs or 60*15/11=5.55*15=82.5 minutes or 1 hr 22 mins approx.