Given:

Time taken by pipe A to fill the tanker = 4 hrs

Time taken by pipe B to fill the tanker = 8 hrs

Time taken by pipe C to fill the tanker = 12 hrs

Time taken by pipe D to empty the tanker = 10 hrs

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Option 1:

Part of the tanker filled in 1 hours when pipes B is open alone

⇒ 1/8

Hence, the tanker will be filled in 8 hours

Option 2:

Part of the tanker filled in 1 hours when pipes A and D are open

⇒ \(\frac{1}{4} - \frac{1}{{10}}{\rm{}} = {\rm{}}\frac{{5 \;-\; 2}}{{20}}{\rm{}} = {\rm{}}\frac{3}{{20}}\)

Hence, the tanker will be filled in \(\frac{{20}}{3}\) hours = 6.6 hours

Option 3:

Part of the tanker filled in 1 hours when pipes A, C and D are open

⇒ \(\frac{1}{4}{\rm{}} + {\rm{}}\frac{1}{{12}} - \frac{1}{{10}}\)

⇒ \(\frac{{15{\rm{\;}} + {\rm{\;}}5 - 6}}{{60}}{\rm{}} = {\rm{}}\frac{{14}}{{40}}{\rm{}} = {\rm{}}\frac{7}{{30}}\)

Hence, the tanker will be filled in \(\frac{{30}}{7}\) hours = 4.3 hours

Option D:

Part of the tanker filled in 1 hours when pipes A, B and D are open

⇒ \(\frac{1}{4}{\rm{\;}} + {\rm{\;}}\frac{1}{8} - \frac{1}{{10}}\)

⇒ \(\frac{{10{\rm{\;}} + {\rm{\;}}5 - 4}}{{40}}{\rm{}} = {\rm{}}\frac{{11}}{{40}}\)

Hence, the tanker will be filled in \(\frac{{40}}{{11}}\) hours = 3.6 hours

∴ Tanker can be filled faster when A, B and D are open.