Let the initial speed of Rahul be $x$ kilometres per hour. Since he travels the distance usually in $6$ hours, (5 pm to 11 pm), the total distance should be $6\times x = 6x$ kilometres.
Let the distance after which Rahul stops for some duration in both scenarios be $y$ kilometres.
In the first scenario, he stops for $20$ minutes or $\dfrac{1}{3}$ hours, therefore, the travel time would be $6-\dfrac{1}{3} = \dfrac{17}{3}$ hours. We have,
$\dfrac{k}{x} + \dfrac{6x-k}{x+3} = \dfrac{17}{3}$
$\Rightarrow \dfrac{6x^2+3k}{x^2+3x} = \dfrac{17}{3}$
$\Rightarrow 18x^2 + 9k = 17x^2 + 51x$
$\Rightarrow x^2 = 51x - 9k$ .....(1)
Similarly, in the second scenario, he stops for $20+10=30$ minutes or $\dfrac{1}{2}$ hours, therefore, the travel time would be $6-\dfrac{1}{2} = \dfrac{11}{2}$ hours. We have,
$\dfrac{k}{x} + \dfrac{6x-k}{x+5} = \dfrac{11}{2}$
$\Rightarrow \dfrac{6x^2 + 5k}{x^2 + 5x} = \dfrac{11}{2}$
$\Rightarrow 12x^2 + 10k = 11x^2 + 55x$
$\Rightarrow x^2 = 55x - 10k$ .....(2)
From equations (1) and (2), we have;
$55x - 10k = 51x - 9k$ or
$4x = k$.
Substituting the value of $k$ in equation (1), we have
$x^2 = 51x - 36x$ or, since $x$ is positive, $x= 15$. Therefore, option B is the correct answer.
