Three pipes A, B & C fill a tank in 3 hours working simultaneously. The pipe C is twice as faster as B and B is twice as fast as A. The time taken by pipe A alone to fill the tank is:

21 hours

The correct answer is Option (4) → 21 hours **

Let the rate of pipe A be $r$.

Then pipe B is twice as fast → rate = $2r$.

Pipe C is twice as fast as B → rate = $4r$.

Together they fill the tank in $3$ hours:

$r + 2r + 4r = \frac{1}{3}$

$7r = \frac{1}{3}$

$r = \frac{1}{21}$

So pipe A alone fills the tank in time:

$T = \frac{1}{r} = 21$ hours

Answer: 21 hours