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Question 413152: Working together three men can paint a barn in six hours. If each man worked alone, the first man would take twice as long as the second man, and the second man would take six hours longer than the third. How many hours would it take the slowest man, working alone, to paint the barn?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Working together three men can paint a barn in six hours. If each man worked alone, the first man would take twice as long as the second man, and the second man would take six hours longer than the third. How many hours would it take the slowest man, working alone, to paint the barn?
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let x=hours third man takes to finish the job alone
work rate=1/x
(x+6)=hours second man takes to finish the job alone
work rate=1/(x+6)
2(x+6)=hours first man takes to finish the job alone
work rate=1/(2(x+6)
6=hours it takes to finish the job working together
work rate=1/6
sum of individual work rates=work rate when working together
(1/x)+(1/(x+6))+(1/(2(x+6))=1/6
LCD=12x(x+6)
12(x+6)+12x+6x=2x(x+6)
12x+72+12x+6x=2x^2+12x
2x^2-18x-72=0
x^2-9x-36=0
(x-12)(x+3)=0
x=12 hours for third man to paint the barn working alone
x=-3 (reject)
ans:
The slowest man on the job is the first man.
He will take 2*(12+6)=36 hours to paint the barn working alone
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