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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

Let's mark work, which must be done as

Productivity of the pipe-A is

Productivity of the pipe-B is

Tank fill rate is:

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

Daric Matthew G. | Daric Matthew G.Daric Matthew G.

What we know:

It fills up faster than it drains

Variables:

X_{f} = amount of liquid in a full tank (units do not matter, pick one)

Equations:

x_{fill}(t)=(X_{f}/9)t

x_{fill}(0)=0

x_{fill}(9)=X_{f}

m=(X_{f}/9)

x_{empt}(t)=(-X_{f}/11)t_{
}

x_{empt}(0)=X_{f}

x_{empt}(11)=0

m=(-X_{f}/11)

Solve:

x_{total}(t)=x_{fill}(t) + x_{empt}(t)

= (X_{f}/9)t + (-X_{f}/11)t

= (11X_{f}/99)t - (9X_{f}/99)t_{
}

= (2X_{f}/99)t

set x_{total}(t)=X_{f}

X_{f} = (2X_{f}/99)t

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)

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**

= (1/9)t + (-1/11)t

= (11/99)t - (9/99)t

= (2/99)t

set xtotal(t)=1

1 = (2/99)t

*Edit

WHOOPS! I forgot to divide

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