SOLUTION: A roofing crew can roof a house in 10 hours. If a second crew was added, the job would take 6 hours. How long would it take the second crew to roof the house alone? Thank you

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A roofing crew can roof a house in 10 hours. If a second crew was added, the job would take 6 hours. How long would it take the second crew to roof the house alone? Thank you      Log On


   



Question 359564: A roofing crew can roof a house in 10 hours. If a second crew was added, the job would take 6 hours. How long would it take the second crew to roof the house alone? Thank you
Answer by CharlesG2(834) About Me  (Show Source):
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A roofing crew can roof a house in 10 hours. If a second crew was added, the job would take 6 hours. How long would it take the second crew to roof the house alone? Thank you


total time T in hours = 1 job/(1 job/F + 1 job/S)
where F is the first roofing crew's time in hours, F = 10 hours
and S = time in hours for the second crew, solve for S
total Time T = 6 hours

6 hours = (1 job)/[(1 job)/(10 hours) + (1 job)/S]
units: hours = job/(job/hours) = job * hours/job = hours

6 = 1/(1/10 + 1/S) (removed units)
6 = 1/(S/10S + 10/10S) (gave each fraction a common denominator)
6 = 1/[(S + 10)/10S]
6 = 10S/(S + 10) (took the reciprocal)
6(S + 10) = 10S
6S + 60 = 10S (distributed the 6)
6S - 10S = -60
-4S = -60 (brought S's to the left, and the 60 to the right)
4S = 60 (divided by -1)
S = 15 hours for the second crew to roof the house alone