It can take 12 hours to …

Suppose the time taken by the pipes of larger and smaller diameters alone to fill the pool be x hours and y hours respectively.

Let the total volume of the pool be V cubic units.
{tex}\therefore{/tex}In 1 hour volume of the water that comes out of the pipe of larger diameter is {tex}\frac { V } { x }{/tex} cubic units.
In 4 hours, the volume of the water that comes out of the pipe of larger diameter is {tex}\frac { 4 V } { x }{/tex} cubic units.
The volume of the water that comes out of the pipe of smaller diameter in {tex}9 \text { hours is } \frac { 9 V } { y }.{/tex}

According to the first condition,
{tex}\therefore \quad \frac { 4 V } { x } + \frac { 9 V } { y } = \frac { 1 } { 2 } V \Rightarrow \frac { 4 } { x } + \frac { 9 } { y } = \frac { 1 } { 2 }{/tex} .............(i)
According to the second condition,
{tex}\therefore \quad \frac { 12 V } { x } + \frac { 12 V } { y } = V \Rightarrow \frac { 12 } { x } + \frac { 12 } { y } = 1{/tex} ............................(ii)
Putting {tex}\frac { 1 } { x } = u \text { and } \frac { 1 } { y } = v{/tex} in (i) and (ii), we obtain
{tex}4 u + 9 v = \frac { 1 } { 2 }{/tex} ..........................(iii)
{tex}12 u + 12 v = 1{/tex} ........................(iv)
Multiplying (iii) by 3 and subtracting from (iv), we get
{tex}- 15 v = - \frac { 1 } { 2 } \Rightarrow v = \frac { 1 } { 30 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 30 } \Rightarrow y = 30{/tex}
Substituting {tex}v = \frac { 1 } { 30 }{/tex} in (iii), we get
{tex}4 u + \frac { 9 } { 30 } = \frac { 1 } { 2 } \Rightarrow 4 u = \frac { 1 } { 2 } - \frac { 9 } { 30 } = \frac { 1 } { 5 } \Rightarrow u = \frac { 1 } { 20 } \Rightarrow \frac { 1 } { x } = \frac { 1 } { 20 } \Rightarrow x = 20{/tex}
Thus, the pipes of larger and smaller diameters fill the swimming pool alone in {tex}20\ hours\ and\ 30 \ hours{/tex} respectively.