If a drain is 1 1/2(3/2) times faster than another, then it drains in 2/3 the time it takes the slower drain.
If one drain is 1 1/2 times faster, the other is 1 1/2 times slower.
For example if a slow drain takes 6 hours, then the faster drain takes (2/3)*6=4 hours.
In other words the slow drain takes 1 1/2 times the fast drain which is (1 1/2)*4=6 hours.
There are different ways to set up the problem.
One solution is the following...
x=# of hours it takes the fast drain
(1.5)x=# of hours it takes the slow drain
Together it takes 8 hours.
(1/x) is the rate per hour of the fast drain
(1/1.5x) is the rate of the slow drain
(1/x)(8)+(1/1.5x)(8)=1 (job finished)
multiply both sides by 3x
24+16=3x (the x's cancel and 3/1.5=2)
x=13 1/3 hours for the fast drain
(1.5)(40/3)=20 hours for the slow drain