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A boy goes to school from home at 6 km/hr and moves back from school at 4 km/hr. If he walks from market to home at the average speed of the previous journey and his father starts walking from his home at the speed of 8 km/hr. When he meets with his father then his father covers 2 km more distance than him. After how many minutes they will meet?

A boy goes to school from home at 6 km/hr and moves back from school at 4 km/hr. If he walks from market to home at the average speed of the previous journey and his father starts walking from his home at the speed of 8 km/hr. When he meets with his father then his father covers 2 km more distance than him. After how many minutes they will meet? Correct Answer 37.5 minutes

Given:

Speed of his father = 8 km/hr

Speed of boy from home to school = 6 km/hr

Speed of boy from school to home = 4 km/hr

Formula:

Average speed = (2 × S1 × S2)/(S1 + S2)

Relative speed = S1 + S2 (when opposite direction)

Distance = speed × time

Calculation:

Average speed of the boy = (2 × 6 × 4) / (10) = 4.8 km/hr

Now,

Difference in distance covered is because one is faster and other is slower.

2 = (8 – 4.8) × time

⇒ time = 2/3.2 = (10/16) × 60 = 75/2 minutes = 37.5 minutes

∴ He and his father will meet on the way after 37.5 minutes. 

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