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?

20

Formula used :-

\(\frac{Speed1 × Speed2 }{Speed \; Difference}\) × Time Difference   = Total Distance

Let Speed of outward journey = x km/h

Speed of return journey = ( x+10 ) km/h

According to question ,

\(\frac{x × (x+10)}{10}\) × 5 = 300

x × (x+10) = 600

On solving ,

x = 20

The speed of the outward journey is 20 km/h .