Explanation:

- Tap A can fill a tank in 10 hours, so it fills 1/10 of the tank in an hour.

- Tap B can fill a tank in 12 hours, so it fills 1/12 of the tank in an hour.

- Both taps start filling the tank at 10 a.m. and need to stop at 4 p.m., which gives a total of 6 hours.

- Together, A and B fill (1/10 + 1/12) per hour. This is 5/60 + 6/60 = 11/60 of the tank per hour.

- In 6 hours, both taps can fill 6 * (11/60) = 66/60 or 1.1 times the tank capacity.

- To fill exactly the tank, reduce the working hours of Tap A. Let x be hours worked by Tap A.

- (x/10) + (6/12) = 1 (complete the tank).

- Solve: (x/10) + 0.5 = 1; x/10 = 0.5; x = 5.

- Close Tap A at 10 a.m. + 5 hours = 3 p.m. and leave B open until 4 p.m.

- Option:2, 3 is correct, as Tap A should close at 3 p.m., and both taps fill the tank completely by 4 p.m.

Option: 2- 3

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