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CAT 2023 Slot 3 QA Question & Solution

ArithmeticHard

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

Options

$12(\sqrt{5} - 2)$
$3(3 + \sqrt{5})$
$3(\sqrt{5} - 1)$
$3(3 - \sqrt{5})$

Solution

Let us assume the speed of the 1st boat is b, the 2nd boat is s, and the river's speed is r.

Let 'd' be the distance between A and B.

=> d = 2(b+r) and d = 3(b-r)

=> b + r = d/2 and b - r = d/3 => r = d/12 (subtracting both equations).

Now, it is given that

$\dfrac{d}{s+r}+\dfrac{d}{s-r}=6$

=> $\dfrac{d}{s+\dfrac{d}{12}}+\dfrac{d}{s-\dfrac{d}{12}}=6$

=> $2ds=6\left(s^2-\dfrac{d^2}{144}\right)$

=> $144s^2-48ds-d^2=0$

Solving the quadratic equation, we get:

$s=d\left(\dfrac{\left(48+\sqrt{\ 48^2+4\left(144\right)}\right)}{2\times\ 144}\right)$

$s=d\left(\dfrac{1}{6}+\dfrac{\sqrt{\ 5}}{12}\right)$

=> Required value of $\dfrac{d}{s+r}$

= $\dfrac{d}{\dfrac{d}{6}+\dfrac{\sqrt{5}d}{12}+\dfrac{d}{12}}$

= $\dfrac{12}{3+\sqrt{\ 5}}=\dfrac{\left(12\right)\left(3-\sqrt{\ 5}\right)}{4}$

= $3\left(3-\sqrt{\ 5}\right)$