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Question 696188: John and Fred hang wallpaper. Working together, they can put up wallpaper in an average sized room in 8 hours. One day, John has an appointment in the morning and Fred is forced to work 5 hours alone. Once John arrives, they work together for 7 hours to complete the room. How long would it take each person to hang wallpaper in an average sized room?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=amount of time it takes John, working alone, to hang the wallpaper
Then John works at the rate of 1/x room per hour
Let y=amount of time it takes Fred, working alone, to hang the wallpaper
Then Fred works at the rate of 1/y of the room per hour
John and Fred, working together, work at the rate of 1/8 room per hour
Soooo
1/x + 1/y =1/8-------------eq1
5*(1/x)+7*(1/8)=1(1 completed room, that is)----eq2
Solve eq 2 for x and substitute into eq1
5/x +7/8=1 multiply each term by 8x
40+7x=8x
x=40 hours-----Time it takes John working alone
From eq 1
1/40+1/y=1/8 multiply each term by 40y
y+40=5y
4y=40
y=10 hours time it takes Fred working alone
CK
1/40 +1/10=1/8
1+4=5
5=5
another way
In 5 hours, John completes (1/40)*5=1/8 of the room
In 7 hours, Both, working together, completes (1/8)*7=7/8 of the room
7/8+1/8=8/8=1 room
Hope this helps---ptaylor
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