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

The Correct Option is C

Let's us assume the speed of the 1^st boat be b, 2^nd 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) : \(3(3-\sqrt5)\).