Pipes A and B together can fill an empty tank in $$6\frac{2}{3}$$ minutes. If A takes 3 minutes more than B to fill the tank, then the time(in minutes) in which A alone would fill one-third part of the tank is:

Let time taken by A to fill an empty tank = $$x$$ min

=> Time taken by B = $$(x-3)$$ min

=> A's efficiency = $$\frac{1}{x}$$ litres/min and B's efficiency = $$\frac{1}{x-3}$$ litres/min

Acc. to ques, => $$[(\frac{1}{x})+(\frac{1}{x-3})]\times(\frac{20}{3})=1$$

=> $$\frac{2x-3}{x^2-3x}=\frac{3}{20}$$

=> $$40x-60=3x^2-9x$$

=> $$3x^2-49x+60=0$$

=> $$x=\frac{49\pm\sqrt{(49)^2-(4\times3\times60)}}{6}$$

=> $$x=\frac{49\pm\sqrt{1681}}{6}$$

=> $$x=\frac{90}{6},\frac{8}{6}$$

$$\because$$ $$x\neq\frac{4}{3}$$

$$\therefore$$ Time taken by A alone to fill one third tank = $$\frac{15}{3}=5$$ minutes

=> Ans - (B)