PSA_PRACTICE_QNR26. A man wants to see his girlfriend who lives 6 miles away. He leaves his house for her house at 6 am and walks at 4 miles an hour. When he leaves, he also sends a pigeon to her house. The bird flies at 30 miles an hour. When the bird reaches her house, she walks toward him at 2 miles an hour. When will they meet on the way?
Let, the man’s house be at point A and his girl-friend’s house be at point B.
Man starts from A and so his pigeon at the same time.
Time taken by the pigeon to reach B= Distance/Speed= 6/(30)=1/5 hr=12 mins.
In 1/5hr, the man travels 4*1/5 miles [ Distance= Speed*Time]
When pigeon reaches B, the man reaches point C.
Total distance to be covered= 26/5 miles
Relative speed= speed of man + speed of pigeon = 4 miles/hr + 2 miles/hr = 6 miles/hr
Time taken to meet= (26/5) / (6) = 26/30 hours= 52 minutes.
Thus, the man and pigeon will meet after 12+52 minutes= 64 minutes. They meet at 6 am + 64 mins= 7:04 am.