Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was

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.