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?

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 ÷ (2/99) = 99/2 = 49.5 (hours)

Answer 4

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

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