Two taps A and B can fill a tank in 10 hours and 12 hours respectively. If the two taps are opened at 10 a.m., then at what time (in p.m.) should the tap A be closed to completely fill the tank at exactly 4 p.m.?

Given, that taps A and B can fill a tank in 10 hours and 12 hours respectively. 

Work done by two taps in 1 hour = ($$\frac{1}{A} + \frac{1}{B}$$) = $$\frac{1}{10} + \frac{1}{12} = \frac{6 + 5}{60} = \frac{11}{60}$$

Let tap A be closed after x hours. Then,

Part filled by (A + B) in x hours + part filled by B in (6 - x) hours = 1

($$\frac{1}{A} + \frac{1}{B})x + \frac{6 - x}{B}$$ = 1

($$\frac{11}{60})x + \frac{6 - x}{12} = 1 \Rightarrow \frac{11x}{60}+\frac{6}{12}-\frac{x}{12}$$ = 1

$$\Rightarrow\frac{6x}{60}+\frac{6}{12}$$ = 1 $$\Rightarrow 6x = 30\Rightarrow x = 5$$

Tap A should be closed after 5 hours i.e at 3 pm

Hence, option B is the correct answer.