A man goes to the fair in Funcity with his son and faithful dog. Unfortunately man misses his son which he realises 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 the son (child) and comes back to the man (father) to show him the direction of his son. It keeps moving to and fro at 60 m/min between son and father, till the man meets the son. what is the distance travelled by the dog in the direction of the son?
Answers
The remaining distance can be found as follows:
Step 1:
In 20 minutes the difference between man and son,
= 20 x 20
= 400 m.
Step 2:
Distance traveled by dog when he goes towards the child,
= 400 x 60/40
= 600 m and time required is 10 minutes.
Step 3:
In 10 min remaining distance between man and child,
= 400 - (20 x 10)
= 200 m.
Time taken by dog to meet the man,
= 200/100
= 2 min
Step 4:
(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 x 20),
= 160 m.
Now, the time taken by dog to meet the child again,
= 160/40
= 4 min.
Step 5:
In 4 min he covers = 4 x 60 = 240 m.
Now, remaining distance in 4 min = 160 - (4 x 20) = 80 m.
Time required by dog to meet the man once again = 80/100 = 0.8 min.
Now remaining distance = 80 - (0.8 x 20) = 64 m.
Answer:can anyone give me the right answer please? I just couldn't solve this
Step-by-step explanation: