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?

Solution (By Examveda Team)

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,
$$\eqalign{ & \Rightarrow \frac{{300}}{x} - \frac{{300}}{{x + 10}} = 5 \cr & \Rightarrow \frac{{300\left( {x + 10} \right) - 300x}}{{x\left( {x + 10} \right)}} = 5 \cr & \Rightarrow 300x + 3000 - 300x = 5x\left( {x + 10} \right) \cr & \Rightarrow 3000 = 5{x^2} + 50x \cr & \Rightarrow 5{x^2} + 50x - 3000 = 0 \cr & \Rightarrow {x^2} + 30x - 20x - 600 = 0 \cr & \Rightarrow x\left( {x + 30} \right) - 20\left( {x + 30} \right) = 0 \cr & \Rightarrow \left( {x - 20} \right)\left( {x + 30} \right) = 0 \cr & \Rightarrow x = 20,\, - 30 \cr} $$
⇒ x = 20 km/hr as negative speed is not possible.
∴ The speed for outward journey is 20 km/hr