It can take $12$ hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for $4$ hours and the pipe of smaller diameter for $9$ hours,only half the pool can be filled. How long would it take for each pipe to fill the pool separately?

(D) Let the time taken by the larger pipe to fill the pool be $x$ hours and the time taken by the smaller pipe be $y$ hours.
In $1$ hour,the larger pipe fills $1/x$ part of the pool and the smaller pipe fills $1/y$ part of the pool.
According to the first condition,if both pipes are used,they fill the pool in $12$ hours:
$1/x + 1/y = 1/12$ --- (Equation $1$)
According to the second condition,the larger pipe works for $4$ hours and the smaller pipe for $9$ hours to fill half the pool:
$4/x + 9/y = 1/2$ --- (Equation $2$)
Let $u = 1/x$ and $v = 1/y$. The equations become:
$u + v = 1/12$ --- (Equation $3$)
$4u + 9v = 1/2$ --- (Equation $4$)
Multiply Equation $3$ by $4$: $4u + 4v = 4/12 = 1/3$.
Subtract this from Equation $4$: $(4u + 9v) - (4u + 4v) = 1/2 - 1/3$.
$5v = 1/6$,so $v = 1/30$.
Substitute $v = 1/30$ into Equation $3$: $u + 1/30 = 1/12$.
$u = 1/12 - 1/30 = (5 - 2)/60 = 3/60 = 1/20$.
Since $u = 1/x = 1/20$,$x = 20$ hours.
Since $v = 1/y = 1/30$,$y = 30$ hours.
Thus,the larger pipe takes $20$ hours and the smaller pipe takes $30$ hours to fill the pool separately.