You can
put this solution on YOUR website!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
You can
put this solution on YOUR website!Two drains takes 9 hours each to completely remove the water from similar-sizes(same volume) pools. The first drain was opened at 6 a.m. Two hours after, the second drain is opened. At what time will the remaining volume in the second pool be three times the remaining volume in the first pool?
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1/9=work rate of both pools
let x=hours drain open for second pool
x+2=hours drain open for first pool
volume=work rate*hours drains are open
x/9=volume drained for second pool
(x+2)/9=volume drained for first pool
..
((x+2)/9)/(x/9)=3
(x+2)/x=3
3x=x+2
2x=2
x=1 hr
Remaining volume in the second pool be three times the remaining volume in the first pool 3 hours after the first drain is open at 9 a.m.
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