Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately.

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 = V

$\therefore$ Volume of the tank filled by the tap of smaller diameter in 1 hour $=\frac{V}{x}$

$\Rightarrow$ Volume of the tank filled by the tap of smaller diameter in 6 hours $=\frac{V}{x} \times 6$

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours $=\frac{V}{(x-9)} \times 6$

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

$\therefore V\left(\frac{1}{x}+\frac{1}{x-9}\right) \times 6=V$

$\Rightarrow \frac{1}{x}+\frac{1}{x-9}=\frac{1}{6}$

$\Rightarrow \frac{x-9+x}{x(x-9)}=\frac{1}{6}$

$\Rightarrow \frac{2 x-9}{x^{2}-9 x}=\frac{1}{6}$

$\Rightarrow 12 x-54=x^{2}-9 x$

$\Rightarrow x^{2}-21 x+54=0$

$\Rightarrow x^{2}-18 x-3 x+54=0$

$\Rightarrow x(x-18)-3(x-18)=0$

$\Rightarrow(x-18)(x-3)=0$

$\Rightarrow x-18=0$ or $x-3=0$

$\Rightarrow x=18$ or $x=3$

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

∴ 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) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.