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

The correct option is:
$3(3-\sqrt{5})$

Let the distance between port A and port B be $d$ km.
Let the speed of the river current be $v$ km/h.
Let the speed of the first boat in still water be $u_1$ km/h.

Step 1: Determine the speeds of the first boat and the river in terms of distance
The first boat takes 2 hours to travel downstream from A to B. Downstream speed is $u_1+v$:
$\frac{d}{u_1+v}=2\Rightarrow u_1+v=\frac{d}{2}$ — (Equation 1)

The first boat takes 3 hours to return upstream from B to A. Upstream speed is $u_1-v$:
$\frac{d}{u_1-v}=3\Rightarrow u_1-v=\frac{d}{3}$ — (Equation 2)

By adding Equation 1 and Equation 2:
$2u_1=\frac{d}{2}+\frac{d}{3}=\frac{5d}{6}\Rightarrow u_1=\frac{5d}{12}$

By subtracting Equation 2 from Equation 1:
$2v=\frac{d}{2}-\frac{d}{3}=\frac{d}{6}\Rightarrow v=\frac{d}{12}$
So, the speed of the first boat in still water is:
$u_1=5v$

Step 2: Determine the speed of the second boat
Let the speed of the second boat in still water be $u_2$ km/h.
The second boat takes a total of 6 hours to travel from port B to port A (upstream) and return to port B (downstream):
$\frac{d}{u_2-v}+\frac{d}{u_2+v}=6$

Substituting $d=12v$:
$\frac{12v}{u_2-v}+\frac{12v}{u_2+v}=6$

Dividing both sides by 6:
$\frac{2v}{u_2-v}+\frac{2v}{u_2+v}=1$

Simplifying the equation:
$2v(u_2+v)+2v(u_2-v)=(u_2-v)(u_2+v)$
$4v{u}_2=u_2^2-v^2$
$u_2^2-4v{u}_2-v^2=0$

Solving for $u_2$ using the quadratic formula:
$u_2=\frac{4v\pm \sqrt{16v^2-4\left(1\right)(-v^2)}}{2}=\frac{4v\pm \sqrt{20v^2}}{2}=(2\pm \sqrt{5})v$

Since speed must be positive and greater than the river speed $v$ to travel upstream:
$u_2=(2+\sqrt{5})v$

Step 3: Compare speeds to identify the slower boat
Speed of the first boat in still water: $u_1=5v$
Speed of the second boat in still water: $u_2=(2+\sqrt{5})v\approx 4.24v$
Since $u_2<u_1$, the second boat is the slower boat.

Step 4: Calculate the time taken by the slower boat to travel from port A to port B
Port A to port B is downstream, so the downstream speed of the slower boat is $u_2+v$:
$\text{Time}=\frac{d}{u_2+v}=\frac{12v}{(2+\sqrt{5})v+v}=\frac{12}{3+\sqrt{5}}$

Rationalizing the denominator:
$\text{Time}=\frac{12(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}=\frac{12(3-\sqrt{5})}{9-5}=\frac{12(3-\sqrt{5})}{4}=3(3-\sqrt{5})$ hours