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 us solve the problem step-by-step by finding the rates of work for each pipe.

Let the total capacity of the tank be represented by the Least Common Multiple (LCM) of the individual times taken by the pipes: 6 hours, 9 hours, and 12 hours.
The LCM of 6, 9, and 12 is 36.
Therefore, let us assume the total capacity of the tank is 36 units.

Now, we calculate the efficiency (work done per hour) of each pipe:

Efficiency of pipe P (filling) = $\frac{36}{6}$ = 6 units/hour.

Efficiency of pipe Q (filling) = $\frac{36}{9}$ = 4 units/hour.

Efficiency of pipe R (emptying) = - $\frac{36}{12}$ = -3 units/hour (negative sign represents emptying the tank).

Step 1: First 4 hours (P and R are open)
Combined efficiency of P and R = 6 - 3 = 3 units/hour.
Water filled in 4 hours = $3\times 4$ = 12 units.

Step 2: Next 6 hours (P is closed, Q and R are open)
Combined efficiency of Q and R = 4 - 3 = 1 unit/hour.
Water filled in these 6 hours = $1\times 6$ = 6 units.

Step 3: Finding the remaining capacity and time taken to fill it
Total water filled so far = 12 + 6 = 18 units.
Remaining capacity of the tank = 36 - 18 = 18 units.
After these 6 hours, pipe R is also closed. Since P was already closed, only pipe Q remains open now.
Efficiency of Q = 4 units/hour.
Time taken by Q to fill the remaining 18 units = $\frac{18}{4}$ = 4.5 hours.

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