Question 1005953
Call the slower pump's time, t.
The faster pump can do it in t-3.
The set up looks like this
2/t + 2/t-3 = 1
Multiply by t(t-3) and get
2(t-3) + 2t = t(t-3)
2t - 6 + 2t = t^2 - 3t
Collecting like terms, then factor and solve...
t^2 - 7t + 6 = 0
(t - 6)(t - 1) = 0
t = 6 or t = 1
but t cannot be one, so 
the slower pump takes 6 hours and the faster takes 3 hours.