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 ->  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


   

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(2198) About Me  (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