Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty in 8 hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?

Height of cone comes down to 50%, => it becomes $$\frac{1}{2}$$

=> Volume would become $$\frac{1}{8}$$ as radius will also become half by similar triangles.

Let the capacity of cone = 24 litres

Volume of water run-off = $$24 - \frac{1}{8} \times 24 = 21$$ litres

Volume of water left in the cone = $$ \frac{1}{8} \times 24 = 3$$ litres

Pipe A's efficiency = $$\frac{24}{8} = 3$$ litres/hr

Pipe B's efficiency = $$\frac{24}{12} = 2$$ litres/hr

Pipe C's efficiency = $$\frac{24}{-4} = -6$$ litres/hr

All will run 19 hours simultaneously (going by the options)

=> Net effect = $$(3 + 2 - 6) \times 19 = -19$$ litres

This means that after 19 hours, 19 litres of water has been removed, we need to remove 2 more litres as per the requirement. Thus,  C will definitely run for another hour.

If we run A and C together for the 20th hour, net effect = $$(3 - 6) \times 1 = -3$$ litres

Run B for 30 minutes => $$2 \times \frac{1}{2} = 1$$ litres

$$\therefore$$ Volume of water removed = $$-19 -3 + 1 = -21$$ litres

Thus, Pipe B was open for 19 hours 30 minutes.