SOLUTION: 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 hous

Question 358990: 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?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
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?
...........
pigeon's speed = 30 mph
distance = 6 miles to her house
time taken by pigeon = 6/30 = 1/5 hour. = 12 minutes
..
speed man walks = 4mph
in 1/5 hour he will walk 4/5 miles.
...
balance distance to her house = 6-4/5
=26/5 miles.
..
man's speed = 4mph
woman's speed = 2mph
they are walking towards each other
so speed = 6 mph.
...
distance = 26/5 miles
speed = 6mph
time = (26/5)/6
=26/5 *1/6
=26/30 hours
= 52 minutes + 12 minutes= 64 minutes
they meet at 7.04 am.
...
m.ananth@hotmail.ca
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