# SOLUTION: An electrician and his apprentice working together, complete the wiring of a building in 20 days. If the apprentice were to do the work on his own he would take 9 days longer than

Algebra ->  Algebra  -> Rate-of-work-word-problems -> SOLUTION: An electrician and his apprentice working together, complete the wiring of a building in 20 days. If the apprentice were to do the work on his own he would take 9 days longer than       Log On

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 Question 142743: An electrician and his apprentice working together, complete the wiring of a building in 20 days. If the apprentice were to do the work on his own he would take 9 days longer than the electrician. How long would it take each of them if they did the work on their own.Answer by ptaylor(2048)   (Show Source): You can put this solution on YOUR website!Let x= amount of time it takes Electrician working alone Then x+9=amount of time it takes the apprentice working alone Electrician works at the rate of 1/x buildings per day Apprentice works at the rate of 1/(x+9) buildings per day Electrician and Apprentice works at the rate of (1/x)+1/(x+9) buildings per day and we are told that the Electrician and Apprentice works at the rate of 1/20 building per day, so: (1/x)+1/(x+9)=1/20 multiply each term by 20x(x+9) 20(x+9)+20x=x(x+9) get rid of parens 20x+180+20x=x^2+9x subtract 40x and also 180 from each side 20x+20x-40x+180-180=x^2+9x-40x-180 or x^2-31x-180=0------------------quadratic in standard form and it can be factored: (x-36)(x+5)=0 Disregard the negative value for x; time in this case is positive x-36=0 x=36 days-----------------------------------------time it takes electrician working alone x+9=36+9=45 days time it takes the apprentice working alone CK 1/36 + 1/45 = 1/20 multiply each term by 180 5+4=9 9=9 Hope this helps---ptaylor