Question

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

• A. $3(\sqrt 5-1)$
• B. $3(3+\sqrt 5)$
• C. $3(3-\sqrt5)$
• D. $12(\sqrt5-2)$

Answer 1

Answer: (C)

$3(3-\sqrt5)$

Answer 2

Let's us assume the speed of the 1st boat be b, 2nd boat be s and the river's speed be r.
Suppose the distance between A and B be d.
⇒ d = 2(b + r) and d = 3(b -r)
⇒ b + r = $\frac{d}{2}$ and b - r = $\frac{d}{3}$
By subtracting both equations, we get :
⇒ r = $\frac{d}{12}$
Given :
$\frac{d}{s+r}+\frac{d}{s-r}=6$

⇒ $\frac{d}{s+\frac{d}{12}}+\frac{d}{s-\frac{d}{12}}=6$

⇒ 2ds = $6(s^2-\frac{d^2}{144})$

⇒ $144s^2-48ds-d^2=0$
By solving the above quadratic equation, we get :
$⇒ s=d(\frac{(48+\sqrt{48^2+4(144)})}{2\times144})$

$⇒s=d(\frac{1}{6}+\frac{\sqrt5}{12})$
Therefore, the required value of $\frac{d}{s+r}$ is as follows :
$=\frac{d}{\frac{d}{6}+\frac{\sqrt5d}{12}+\frac{d}{12}}$

$=\frac{12}{3+\sqrt5}$

$=\frac{(12)(3-\sqrt5)}{4}$

$=3(3-\sqrt5)$

Therefore, the correct option is (C) : $3(3-\sqrt5)$.