SOLUTION: solve. Round to the nearest tenth of necessary Mr. Endicott, Mr. Bookout, and Mr. Piatt are going to paint a house together. Mr. Endicott can paint one side of the house in 4

Question 1034412: solve. Round to the nearest tenth of necessary

Mr. Endicott, Mr. Bookout, and Mr. Piatt are going to paint a house together. Mr. Endicott can paint one side of the house in 4 hrs. to paint an equal area, Mr. Bookout takes 3 hrs and Mr. Piatt takes 2 hrs. If the men work together, how long will it take them to paint one side of the house?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
e = the rate that mr. endicott can paint the side of a house.
b = the rate that mr. bookout can paint the side of a house.
p = the rate that mr. piatt can paint the side of a house.

e = 1/4
b = 1/3
p = 1/2

this is derived from the general formula of r * t = q
r is the rate of work.
t is the time.
q is the quantity of work.

q = 1 side of a house.
t = number of hours to paint the side of the house.
r is the rate at which the side of the house is painted.

for mr. endicott, the formula becomes r * 4 = 1
solve for r to get 1/4.

do the same for mr. bookman and mr. piatt and you get the rates shown above.

when they work together, their rates are additive.

you get (r1 + r2 + r2) * t = q
q = 1
r1 = e = 1/4
r2 = b = 1/3
r3 = p = 1/2

you get (1/4 + 1/3 + 1/2) * t = 1

least common denominator looks to be 12.

you get (3/12 + 4/12 + 6/12) * t = 1

combine like terms to get 13/12 * t = 1

divide both sides of the equation by 13/12 to get t = 1 / (13/12)

this is the same as t = 1 * (12/13) = 12/13

it would take them 12/13 of an hour to paint the side of the house if they all work together.

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