A tank is filled in 4 hours by three pipes A, B and C. The pipe C is twice as fast as B and pipe B is thrice as fast as A. How much time will pipe A alone take to fill the tank?

let A takes 6x hours to fill the tank.

B takes 2x hours.

C takes x hours.

tank filled by three taps in 1 hour $$=\dfrac{1}{x}+\dfrac{1}{2x}+\dfrac{1}{6x}$$    

tank filled by three taps in  hour $$=4(\dfrac{1}{x}+\dfrac{1}{2x}+\dfrac{1}{6x})$$

according to problem,

$$\rightarrow 4(\dfrac{1}{x}+\dfrac{1}{2x}+\dfrac{1}{6x})=1$$

$$4(\dfrac{6+3+1}{6x})=1$$

$$4(\dfrac{10}{6x})=1$$

$$6x=40$$

time taken by tap A=6x=40hours