Direction: The line graph shows the time (in hours) taken by two pipes A and B to fill...

Read Values from Graph for Tank R**:

• Pipe A fills Tank R in $18\text{ hours} \implies R_A = \frac{1}{18}\text{ of tank/hour}$.
• Pipe B fills Tank R in $24\text{ hours} \implies R_B = \frac{1}{24}\text{ of tank/hour}$.
• Pipe D fills Tank R in $36\text{ hours} \implies R_D = \frac{1}{36}\text{ of tank/hour}$.

2. Normalize Work Units:

   • $\text{LCM}(18, 24, 36) = 72\text{ units}$.
   • Rate of Pipe A = $72 / 18 = 4\text{ units/hour}$.
   • Rate of Pipe B = $72 / 24 = 3\text{ units/hour}$.
   • Rate of Pipe D = $72 / 36 = 2\text{ units/hour}$.

3. Operational Schedule (2-Hour Cycle):

   • 'Pipes B and D are opened for one hour each, alternatively with pipe A' means:
     • Hour 1: Pipe A + Pipe B $\implies 4 + 3 = 7\text{ units}$.
     • Hour 2: Pipe A + Pipe D $\implies 4 + 2 = 6\text{ units}$.
   • Work done in one complete $2\text{-hour}$ cycle = $7 + 6 = 13\text{ units}$.

4. Calculate Total Duration:

   • In $5$ cycles ($10\text{ hours}$): $5 \times 13 = 65\text{ units}$ are filled.
   • Remaining volume to fill = $72 - 65 = 7\text{ units}$.
   • In the $11\text{th hour}$, it is the turn of $(A + B)$, which fills exactly $7\text{ units/hour}$:
     $$\text{Time for remaining 7 units} = \frac{7}{7} = 1\text{ hour}$$
   • Total time required = $10 + 1 = \mathbf{11\text{ hours}}$.

Thus, the correct option is C.