Two pipes A and B can fill a tank in $$3\frac{\text{3}}{\text{7}}$$ hours when opened simultaneously. If B alone can take 2 hours less than A alone takes to fill the tank completely. How much time does A alone take to fill the tank?

Let us assume that A alone takes 'x' hours to fill the tank.

B alone takes 'x - 2' hours to fill the tank.

Part of the tank filled by pipe A in 1 hour = $$\frac{1}{x}$$

Part of the tank filled by pipe B in 1 hour = $$\frac{1}{x-2}$$

Together they fill the tank in $$3\frac{3}{7}$$ hr = $$\frac{24}{7}$$ hr

$$\frac{24}{7}(\frac{1}{x} + \frac{1}{x-2}) = 1$$

On simplifying, we get

$$24(2x-2) = 7x(x-2)$$

$$7x^2-62x+48 = 0$$

$$(x-8)(x-\frac{6}{7})=0$$

$$\therefore x = 8$$ ( $$x$$ can't take the value of $$\frac{6}{7}$$ as $$x-2$$ is negative )

Hence, A alone takes 8 hr to fill the tank completely.