Question

A person travelled a distance of 60 km and then returned to the starting point. The time taken by him for the return journey was $\frac{1}{2}$ hour more than the time taken for the outward journey, and the speed during the return journey was 10 km/h less than that during the outward journey. His speed during the outward journey (in km/h) was:

• A. 24
• B. 40
• C. 36
• D. 30

Answer 1

Answer:

B. 40

Solution:

Given :

The person traveled 60 km and returned to the same point

Time taken for return journey = 1/2 Hr + Time taken for outward journey

Speed during return journey = Speed during the outward journey - 10 km/h 

Formula used :

Speed = Distance/Time

Calculation :

Let the Speed of outward journey be S₁ and speed for return journey be S₂

Also, Let the time taken for outward journey be T₁ and time taken for return journey be T₂

 According to question,

S₁ = 60/ T₁ 

⇒ T₁ = 60/S₁      ----(1)

Also, from given information in question

T₁ + 1/2 = T₂     ----(2)

S₂ = S₁ - 10      ----(3)

For return journey

T₂ = 60 / S₂

From Eq (2) and (3), we have

⇒ T₁ + 1/2 = 60 / (S₁ - 10)

⇒ (60/S₁) +1/2 = 60 / (S₁ - 10)

⇒ 600/(S1−10)S1=1/2

⇒ S₁ (S₁ - 10) = 1200

⇒ S₁ = 40

∴ Speed of the person during outward journey is 40 km/h