Two pipes P and Q can fill a tank in 6 hours and 9 hours respectively, while a third pipe R can empty the tank in 12 hours. Initially, P and R are open for 4 hours. Then P is closed and Q is opened. After 6 more hours R is closed. The total time taken to fill the tank (in hours) is _________.

The correct option is 14.50.

Let's break down the solution step-by-step to understand why this is the correct answer.

Step 1: Determine the capacities and rates of the pipes
Let the total capacity of the tank be represented by the Least Common Multiple (LCM) of the times taken by the individual pipes: 6 hours, 9 hours, and 12 hours.
LCM(6, 9, 12) = 36 units.
Therefore, let the total capacity of the tank be 36 units.

Now, we calculate the hourly rate of work for each pipe:
Rate of pipe P (filling) = $\frac{36}{6}$ = +6 units/hour.
Rate of pipe Q (filling) = $\frac{36}{9}$ = +4 units/hour.
Rate of pipe R (emptying) = $\frac{-36}{12}$ = -3 units/hour (negative because it empties the tank).

Step 2: Calculate work done in the first 4 hours
Initially, pipes P and R are opened together for 4 hours.
Combined rate of P and R = 6 - 3 = +3 units/hour.
Work done in 4 hours = $3\times 4$ = 12 units.

Step 3: Calculate work done in the next 6 hours
After 4 hours, P is closed and Q is opened. R remains open. This state lasts for 6 hours.
Combined rate of Q and R = 4 - 3 = +1 unit/hour.
Work done in these 6 hours = $1\times 6$ = 6 units.

Step 4: Calculate the remaining work and time needed to finish
Total work completed so far = 12 + 6 = 18 units.
Remaining work to fill the tank = 36 - 18 = 18 units.
After this period (total of 4 + 6 = 10 hours elapsed), pipe R is closed. Now, only pipe Q is open.
Rate of pipe Q = 4 units/hour.
Time taken by Q to fill the remaining 18 units = $\frac{18}{4}$ = 4.5 hours.

Step 5: Calculate total time taken
Total time taken = Time for (P + R) + Time for (Q + R) + Time for Q alone
Total time = 4 hours + 6 hours + 4.5 hours = 14.50 hours.