Three pipes, D, E and F, can fill a tank in 6 min, 8 min and 12 min, respectively. All the pipes are opened simultaneously, and then pipes D and E are closed 3 minutes before tank is full. In how much time will the tank be full?

Let's assume the tank will be full in 'y' min.

Three pipes, D, E and F, can fill a tank in 6 min, 8 min and 12 min, respectively.

LCM of 6, 8 and 12 is 24 units which can be assumed as the total capacity of the tank.

Efficiency of pipe D = $$\frac{24}{6}$$ = 4 units/min

Efficiency of pipe E = $$\frac{24}{8}$$ = 3 units/min

Efficiency of pipe F = $$\frac{24}{12}$$ = 2 units/min

All the pipes are opened simultaneously, and then pipes D and E are closed 3 minutes before tank is full.

$$4\times(y-3)+3\times(y-3)+2\times y = 24$$

$$4y-12+3y-9+2y=24$$

9y-21 = 24

9y = 24+21

9y = 45

y = 5

So the tank will be full in 5 min.