A man goes to the fair with his son and dog. Unfortunately man misses his son which he realizes 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to son and comes back to the man to show him the direction of his son. He keeps moving to and fro at 60 m/min between son and father, till the man meets the son. What is the distance traveled by the dog in the direction of the son?

Solution(By Examveda Team)

In 20 minutes the difference between man and son,
= 20 × 20
= 400 m.
Distance traveled by dog when he goes towards the child,
= $$400 \times \frac{{60}}{{40}}$$
= 600 m and time required is 10 minutes.
In 10 min remaining distance between man and child,
= 400 - (20 × 10)
= 200 m.
Time taken by dog to meet the man,
= $$\frac{{200}}{{100}}$$
= 2 min

(Relative speed of dog with child is 40 m/min and same with man is 100 m/min.)

Remaining distance in 2 min,
= 200 - (2 × 20),
= 160 m.
Now, the time taken by dog to meet the child again,
= $$\frac{{160}}{{40}}$$
= 4 min.
In 4 min he covers = 4 × 60 = 240 m.
Now, remaining distance in 4 min = 160 - (4 × 20) = 80 m.
Time required by dog to meet the man once again = $$\frac{{80}}{{100}}$$ = 0.8 min.
Now remaining distance = 80 - (0.8 × 20) = 64 m.