Two pipes can fill a tank in 6 hour. The large pipe works twice as fast as the smaller pipe. How long would it take each pipe to fill the tank if they worked separately ?

Let x be the time it takes for the smaller pipe to fill the tank.
Therefore, the larger pipe takes x/2 time to fill the tank.
The rate of the smaller pipe is 1/x of the tank per hour.
The rate of the larger pipe is 1/(x/2) = 2/x of the tank per hour.
Together, the rate of both pipes is 1/x + 2/x = 3/x of the tank per hour.
Since they can fill the tank in 6 hours, their combined rate is 1/6 of the tank per hour.
Therefore, 3/x = 1/6.
Multiplying both sides by 6x, we get 18 = x.
Thus, it takes the smaller pipe 18 hours to fill the tank.
The larger pipe takes 18/2 = <<18/2=9>>9 hours to fill the tank. Answer: \boxed{18, 9}.