Question 293528
It takes Larry 8 hours to paint his house.
Larry paints for 3 hours by himself.
His friend Patrick joins him and they finish in 2 hours.
How long would it take Patrick to paint the house himself?


Larry would take 8 hours to paint the house by himself.


His rate of painting is 1/8 of the house per hour.


This is based on the formula R * T = U


The number of units is equal to R is equal to 1.


The time is equal to 8 hours.


Plug those numbers into the equation to get R * 8 = 1


Divide both sides of that equation by 8 to get R = 1/8.


Plug 1/8 into that formula to get 1/8 * 8 = 1 so this rate is good.


in 3 hours Larry has painted 3/8 of the house.


That leaves 5/8 of the house left to do.


Larry and Patrick work together to finish the rest of the house in 2 hours.


Since they are working together their rates are additive.


Larry's rate is 1/8 of the house per hour.


Patrick's rate of painting the house per hour is represented by x.


The formula is Number of Units = Rate per Unit * Time


The number of units is equal to 1 if they had to do the whole house.


Since they only have to do 5/8 of the house, then the number of units is 5/8.


Since they are working together, their rates are combined.


The combines rate is (1/8 + x)


the amount of time they took is 2 hours.


The formula becomes:


5/8 = (1/8 + x) * 2


Simplify by removing parentheses to get:


5/8 = 2/8 + 2*x


Subtract 2/8 from both sides of the equation to get:


5/8 - 2/8 = 2*x


Combine like terms to get:


3/8 = 2*x


Divide both sides of the equation by 2 to get:


3/16 = x


Since x is Patrick's rate, this means that Patrick can paint 3/16 of the house in 1 hour.


This means that the formula for Patrick to do the whole house himself would be:


R * T = U where R = 3/16 and U = 1.


This formula becomes (3/16) * T = 1


Divide both sides of this equation by 3/16 to get T = 1/(3/16) which is the same as T = 16/3 which simplifies to T = 5 and 1/3 hours.


It would take Patrick 5 and 1/3 hours to paint the house by himself.


The key to this type of problem is the number of units that are left to be done.