A boat takes 2 hours to travel downstream a river from port A to port B, and 3 hours to return to port A. Another boat takes a total of 6 hours to travel from port B to port A and return to port B. If the speeds of the boats and the river are constant, then the time, in hours, taken by the slower boat to travel from port A to port B is

Let us assume the speed of the 1st boat is b, the 2nd boat is s, and the river's speed is r.

Let 'd' be the distance between A and B.

=> d = 2(b+r) and d = 3(b-r)

=> b + r = d/2 and b - r = d/3 => r = d/12 (subtracting both equations).

Now, it is given that

$$\dfrac{d}{s+r}+\dfrac{d}{s-r}=6$$

=> $$\dfrac{d}{s+\dfrac{d}{12}}+\dfrac{d}{s-\dfrac{d}{12}}=6$$

=> $$2ds=6\left(s^2-\dfrac{d^2}{144}\right)$$

=> $$144s^2-48ds-d^2=0$$

Solving the quadratic equation, we get:

$$s=d\left(\dfrac{\left(48+\sqrt{\ 48^2+4\left(144\right)}\right)}{2\times\ 144}\right)$$

$$s=d\left(\dfrac{1}{6}+\dfrac{\sqrt{\ 5}}{12}\right)$$

=> Required value of $$\dfrac{d}{s+r}$$

= $$\dfrac{d}{\dfrac{d}{6}+\dfrac{\sqrt{5}d}{12}+\dfrac{d}{12}}$$

= $$\dfrac{12}{3+\sqrt{\ 5}}=\dfrac{\left(12\right)\left(3-\sqrt{\ 5}\right)}{4}$$

= $$3\left(3-\sqrt{\ 5}\right)$$