First find the pipe rates. Let pipe B alone need x hours, so pipe A alone needs x − 5 hours. Together they fill the reactor in 6 hours, so 1/(x − 5) + 1/x = 1/6. Combining the left side gives (2x − 5)/(x² − 5x) = 1/6, so 6(2x − 5) = x² − 5x, which simplifies to x² − 17x + 30 = 0, or (x − 15)(x − 2) = 0. The root x = 2 is rejected because it would make pipe A's time negative. So pipe B alone takes 15 hours and pipe A alone takes 10 hours.
Now convert to volumes for clean arithmetic. Take the reactor as 360 liters. Pipe A fills 360 ÷ 10 = 36 liters per hour; pipe B fills 360 ÷ 15 = 24 liters per hour; the drain empties 360 ÷ 12 = 30 liters per hour.
The first stage fills the reactor to half full, leaving 180 liters to go. From the moment pipe B is shut off and the drain opens, only pipe A runs against the drain, so the net rate is 36 − 30 = 6 liters per hour. The time to add the last 180 liters is 180 ÷ 6 = 30 hours. The answer is D.
Each wrong choice drops one feature of the staged finish. Choice A (5) uses pipe A's gross rate of 36 liters per hour and ignores that the drain is open. Choice B (6) leaves pipe B running against the drain instead of shutting it off, giving a net rate of 36 + 24 − 30 = 30 liters per hour and 180 ÷ 30 = 6. Choice C (15) is pipe B's alone time, an intermediate quantity someone reports instead of continuing to the staged final-stage duration the question asks for. Choice E (60) uses the full 360 liters and forgets that the first stage already filled half the reactor, so only 180 liters remain when the drain opens.