Let the tap of smaller diameter fill the tank in x hours.
∴ Time taken by the tap of larger diameter to fill the tank = (x - 9)h
Suppose the volume of the tank be V.
Volume of the tank filled by the tap of smaller diameter in x hours = F
∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = F/x
⇒ Volume of the tank filled by the tap of smaller diameter in 6 hour = F/x x 6
Similarly
Volume of the tank filled by the tap of larger diameter in 6 hours = F/(x - 9) x 6
Now,
Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possible.
∴ x = 18
Time taken by the tap of smaller diameter to fill the tank = 18 h
Time taken by the tap of larger diameter to fill the tank = (18 - 9) = 9h
Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.