Posted by clara on .
lean can paint her new apartment 3 hours longer than his friend,carlos.if both can do painting in 2 hours,how long will it take each one to finish the job alone?

algebra 
Bayarbold,
labour*time=job time=job/labour
time=3 and 3x=1 x=1/3
If both can do painting in 2 hours,So
(x+y)2=1 x=1/3 hence (1/3+y)*2=1
And y is 1/6.We can say that it take 6hours to finish the job alone. 
algebra 
Reiny,
let Carlos' time be x hrs
so rate for Carlos = 1/x
Lean's time is x+3
Lean's rate = 1/(x+3)
combined rate = 1/x + 1/(x+3)
= (x+3 + x)/(x(x+3)) = (2x+3)/(x(x+3))
1/[ (2x+3)/(x(x+3))] = 2
x(x+3)/(2x+3) = 2
x^2 + 3x = 4x + 6
x^2  x  6 = 0
(x3)(x+2) = 0
x = 3 or x=2, the latter is not possible
Carlos would take 3 hrs, and Lean would take 6 hrs alone
check:
combined rate = 1/3 + 1/6 = 1/2
combined time = 1/(1/2) = 2 
algebra 
Sharon,
Keiko bicycles 4km/h faster than Carlos. In the same time it takes Carlos to Bicycle 45km, Keiko can bicycle 57km. How fast does each bicyclist travel?