A person covers a distance of 300 km and then returns to the starting point. The time taken by him for the outward journey is 5 hours more than the time taken for the return journey. If he returned at a speed of 10 km/h more than the speed of going. What was the speed (in km/h) for the outward journey?

Answer: Option D

Let the speed of outward journey be $$x$$ km/hr
Hence, the speed of return journey will be ($$x$$ + 10) km/hr
So, time taken in return journey = $$\frac{300}{x+10}\text{hr}$$
Time taken in outward journey = $$\frac{300}{x}\text{hr}$$
According to the question,
$$\begin{array}{rl}& \Rightarrow \frac{300}{x}-\frac{300}{x+10}=5\\ & \Rightarrow \frac{300\left(x+10\right)-300x}{x\left(x+10\right)}=5\\ & \Rightarrow 300x+3000-300x=5x\left(x+10\right)\\ & \Rightarrow 3000=5x^2+50x\\ & \Rightarrow 5x^2+50x-3000=0\\ & \Rightarrow x^2+30x-20x-600=0\\ & \Rightarrow x\left(x+30\right)-20\left(x+30\right)=0\\ & \Rightarrow \left(x-20\right)\left(x+30\right)=0\\ & \Rightarrow x=20,-30\end{array}$$
⇒ x = 20 km/hr as negative speed is not possible.
∴ The speed for outward journey is 20 km/hr