Question 621623
Each drain empties at the rate of (1/9)V per hour where V is the volume of each pool

When the second drain is opened, the first pool has emptied 2*(1/9)V leaving V-(2/9)V=(7/9)V yet to be emptied.

Let t=number of hours it takes AFTER the second drain is opened for the second pool to have three time the remaining volume of the first pool

Amount of water remaining in the first pool after t hours=(7/9)V-(t/9)V
Amount of water remaining in the second pool after t hours V-(t/9)V and this is three time the remaining volume of the first pool, soooo:
3{(7/9)V-(t/9)V}=V-(t/9)V simplify
(21/9)V-(3t/9)V=V-(t/9)V
(21/9)V-V=(3t/9)V-(t/9)V
(12/9)V=(2t/9)V divide each side by V and multiply each side by 9
2t=12
t=6 hr--number of hours it takes AFTER the second drain is opened
Second drain is opened at 8 am; 8 am plus 6 hours=2 pm  --the time at which the second pool has 3 times the volume of the first pool

CK
In 6 hours the first pool has drained 6+2=8/9 its volume, leaving (1/9)V
In 6 hours the second pool has drained 6/9 its volume leaving (3/9)V and this is three times more that (1/9)V

Hope this helps---ptaylor