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## A tank can be filled in 9 hours and drained in 11 hours. How long will it take to fill if the drain is left open?

A tank can be filled in 9 hours and drained in 11 hours. How long will it take to fill if the drain is left open?

rate *time=output
t=time
rate of fill *t=Volume of tank (filled)
rate of fill*9=V
rate of fill=V/9

rate of drain*t=Volume of tank(drained)
rate of drain*11=V
rate of drain=V/11

time needed to fill the tank with both drains open:
[(V/9)-(V/11)]*t=V
[(1/9)-(1/11)]*t=1
(2/99)*t=1
t=99/2 or 49.5 hrs

check: (99/2)*(1/9)=11/2=5 1/2 tanks filled
(99/2)*(1/11)=9/2=4 1/2 tanks drained
difference is 1 tank filled

Excellent explanation.
Thank you, Jason, for the kind words. It was greatly appreciated.
After 9 hours 9/11 of the tank will be drained, so 2/11 will be filled. Since you need to fill 1 tank, you will have to repeat 9-hour cycle 11/2=5.5 times, that is, total time required to fill the tank is 9*11/2=49.5 hours.
Let's assume that pipe-A fills the tank, and pipe-B drains the tank.

Let's mark work, which must be done as "1"

Productivity of the pipe-A is 1/9 (unit per hour)

Productivity of the pipe-B is 1/11 (unit per hour)

Tank fill rate is:

(1/9) - (1/11) = 2/99

1 รท (2/99) = 99/2 = 49.5 (hours)
What we know:
It fills up faster than it drains

Variables:
Xf = amount of liquid in a full tank (units do not matter, pick one)

Equations:
xfill(t)=(Xf/9)t
xfill(0)=0
xfill(9)=Xf
m=(Xf/9)
xempt(t)=(-Xf/11)t
xempt(0)=Xf
xempt(11)=0
m=(-Xf/11)

Solve:
xtotal(t)=xfill(t) + xempt(t)
= (Xf/9)t + (-Xf/11)t
= (11Xf/99)t - (9Xf/99)t
= (2Xf/99)t
set xtotal(t)=Xf
Xf = (2Xf/99)t
t=99/2

Check:
Choose full tank = 1
xfill(t)=(1/9)t
xfill(0)=0
xfill(9)=1
m=(1/9)
xempt(t)=(-1/11)t
xempt(0)=1
xempt(11)=0
m=(-1/11)

xtotal(t)=xfill(t) + xempt(t)
= (1/9)t + (-1/11)t
= (11/99)t - (9/99)t
= (2/99)t
set xtotal(t)=1
1 = (2/99)t
t=99/2
*Edit
WHOOPS! I forgot to divide