Boats and Streams with Multiple Boats

Let distance $AB = D$ and river speed = $v$.

Boat 1: Downstream speed $b_1 + v = D/2$ Upstream speed $b_1 - v = D/3$ Subtracting gives $2v = D/2 - D/3 = D/6 \implies v = D/12$.

Boat 2: Takes 6 hours for round trip (upstream B to A + downstream A to B): $\frac{D}{b_2 - v} + \frac{D}{b_2 + v} = 6$ Let $b_2 = k \cdot \frac{D}{12}$. Substituting $v = D/12$: $\frac{12}{k-1} + \frac{12}{k+1} = 6 \implies \frac{2}{k-1} + \frac{2}{k+1} = 1$ $2(k+1) + 2(k-1) = k^2 - 1 \implies k^2 - 4k - 1 = 0$ Since $k > 1$, $k = 2 + \sqrt{5}$. So $b_2 = (2 + \sqrt{5})\frac{D}{12} \approx 4.236 \frac{D}{12}$.

Speed of Boat 1: $b_1 = 5\frac{D}{12}$. Thus, Boat 2 is the slower boat.

Time taken by Boat 2 to travel from A to B (downstream): $\text{Speed downstream} = b_2 + v = (2 + \sqrt{5})\frac{D}{12} + \frac{D}{12} = (3 + \sqrt{5})\frac{D}{12}$ $\text{Time} = \frac{D}{(3 + \sqrt{5})D/12} = \frac{12}{3 + \sqrt{5}} = \frac{12(3 - \sqrt{5})}{9 - 5} = 3(3 - \sqrt{5})$