A person travels a distance of 300 km to his destination and then returns to the starting point. The time taken by him for the onward journey towards his destination is 5 hours more than the time taken for the return journey towards the starting point. If he returns at a speed of 10 km/hr more than the onward speed, what is his average speed (in km/h) for the entire journey?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 24
Distance covered in the onward journey = 300 km = Distance covered in the return journey
Let's assume his onward speed = x km/hr
Return speed of the person = (x + 10) km/hr
Time taken to reach the destination = 300/x
Time taken to return to the starting point = 300/x + 10
Now, according to the question,
(300/x) - 300/(x + 10) = 5
⇒ (300x + 3000 - 300x)/(x2 + 10x) = 5
⇒ 3000 = 5x2 + 50x
⇒ x2 + 10x - 600 = 0
⇒ x2 + 30x - 20x - 600 = 0
⇒ (x + 30)(x - 20) = 0
⇒ x = -30 can't be possible
⇒ x = 20 km/h
Onward speed x = 20 km/h
Return speed = 20 + 10 = 30 km/h
Average speed = Total distance/total time
⇒ Average speed = 600/(300/20 + 300/30)
⇒ 600/(15 + 10)
⇒ 24 km/h