A person travels 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 returns at a speed of 10 km/h more than the speed of going, what is the average speed (in km/h)for the entire journey?

24

Let speed of outward journey = S km/h

Speed while returning = (S + 10) km/h

According to question ,

\(\frac{300}{S}\) - \(\frac{300}{(S+10)}\) = 5

\(\frac{300(S+10) - 300x}{S(S+10)}\) = 5

3000 = 5S (S+10)

On solving ,

S = 20 km/h

Time of Outward journey ,

Time = \(\frac{Distance}{Speed}\)

= \(\frac{300}{20}\) = 15 hours

Time for return journey = \(\frac{300}{20 + 10 }\)

= \(\frac{300}{30 }\) = 10 hours

Average speed = \(\frac{300 + 300}{15 + 10 }\)

= \(\frac{600}{25  }\)

= 24 km/h