Question

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?

• A. 15
• B. 24
• C. 20
• D. 30

Answer 1

Answer:

B. 24

Solution:

Given:

Distance travelled = 300 km

Time is taken by outward journey = Time taken for return journey + 5 hrs

Speed of returning journey = Speed of going journey + 10 km/h

Concept used:

Average speed = (Total distance traveled)/(Total time is taken)

Calculation:

Let the speed of the outward journey be x km/h.

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

Time is taken by outward journey = 300/x hrs

Time is taken by returning journey = 300/(x + 10) hrs

According to the question, 

Time is taken by outward journey = Time taken for return journey + 5 hrs

⇒ 300/x = 300/(x + 10) + 5

⇒ 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(x + 30) – 20(x + 30) = 0

⇒ (x + 30)(x – 20) = 0

⇒ (x + 30) = 0 or (x – 20) = 0

⇒ x = -30 or x = 20

Speed of outward journey = 20 hrs

Time is taken by outward journey = 300/x hrs

⇒ Outward journey time = 300/20

⇒ Outward journey time = 15 hrs

Time is taken by outward journey = Time taken for return journey + 5 hrs

⇒ 15 = Returning time + 5

⇒ Returning time = 10 hrs

Average speed = (Total distance traveled)/(Total time is taken)

⇒ Average speed = (300 + 300)/(15 + 10)

⇒ Average speed = 600/25

⇒ Average speed = 24 km/h

∴ The average speed for the entire journey is 24 km/h.