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

Think of this like two boats racing on a river that has flowing water. The flowing water helps boats go one way (downstream) but makes it harder to go the other way (upstream).

When boats go WITH the current they go faster. When boats go AGAINST the current they go slower.

---

Let's call:

• $v_1 =$ speed of Boat $1$ in calm water (no current)
• $v_2 =$ speed of Boat $2$ in calm water (no current)
• $c =$ speed of river current

Remember:

• Downstream speed $=$ boat speed $+$ current speed
• Upstream speed $=$ boat speed $-$ current speed

---

Boat $1$ takes $2$ hours $A \to B$ and $3$ hours $B \to A$

Since Distance $=$ Speed $\times$ Time, and the distance $A$ to $B$ equals distance $B$ to $A$:

Going downstream $(A \to B)$: Distance $= (v_1 + c) \times 2$

Going upstream $(B \to A)$: Distance $= (v_1 - c) \times 3$

These distances must be equal:

$(v_1 + c) \times 2 = (v_1 - c) \times 3$

Solving:

$2v_1 + 2c = 3v_1 - 3c$

$2c + 3c = 3v_1 - 2v_1$

$5c = v_1$

Boat $1$'s speed in still water is $5$ times the current speed.

---

Using $v_1 = 5c$:

• Downstream speed $= v_1 + c = 5c + c = 6c$
• Distance $= 6c \times 2 = 12c$

For easy calculation, let's set $c = 1$:

• Current speed $= 1$ unit
• Boat $1$ speed $= 5$ units
• Distance between ports $= 12$ units

---

Boat $2$ takes total $6$ hours for $B \to A \to B$

Journey breakdown:

• $B$ to $A$ (upstream): Time $= \dfrac{12}{v_2 - 1}$
• $A$ to $B$ (downstream): Time $= \dfrac{12}{v_2 + 1}$

Total time equation:

$\dfrac{12}{v_2 - 1} + \dfrac{12}{v_2 + 1} = 6$

---

Multiply everything by $(v_2 - 1)(v_2 + 1)$ to clear fractions:

$12(v_2 + 1) + 12(v_2 - 1) = 6(v_2 - 1)(v_2 + 1)$

$12v_2 + 12 + 12v_2 - 12 = 6(v_2^2 - 1)$

$24v_2 = 6v_2^2 - 6$

$6v_2^2 - 24v_2 - 6 = 0$

Divide by $6$: $v_2^2 - 4v_2 - 1 = 0$

---

For $v_2^2 - 4v_2 - 1 = 0$:

$v_2 = \dfrac{4 \pm \sqrt{16 + 4}}{2} = \dfrac{4 \pm \sqrt{20}}{2} = \dfrac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$

Since speed must be positive and greater than current $(1)$:

$v_2 = 2 + \sqrt{5} \approx 4.236$

---

Boat speeds in still water:

• Boat $1$: $5$ units
• Boat $2$: $2 + \sqrt{5} \approx 4.236$ units

Boat $2$ is slower.

---

Boat $2$ going downstream $(A \to B)$:

• Speed $= (2 + \sqrt{5}) + 1 = 3 + \sqrt{5}$
• Time $= \dfrac{12}{3 + \sqrt{5}}$

Rationalize the denominator (multiply top and bottom by conjugate):

$\dfrac{12}{3 + \sqrt{5}} \times \dfrac{3 - \sqrt{5}}{3 - \sqrt{5}} = \dfrac{12(3 - \sqrt{5})}{9 - 5} = \dfrac{12(3 - \sqrt{5})}{4} = 3(3 - \sqrt{5})$

Therefore, the time for the slower boat to travel from $A$ to $B$ is $9 - 3\sqrt{5}$ hours.