Fred begins walking toward John house at 3 mi/h. John leaves his house at the same time and walks toward Fred's house on the sam
e path at a rate of 2 mi/h. How long will it be before they meet if the distance between the houses is 4 miles?
1 answer:
Let us assume as at when they meet, the time elapsed is x hours.
Fred 3 mi/h
John 2 mi/h
Distance = speed * time
Fred = 3x
John = 2x
Houses are 4miles apart.
So: 3x + 2x = 4 5x = 4
x = 4/5 hour = 0.8 hour or 0.8*60 = 48 minutes
Time elapsed is 0.8 hour or 48 minutes.
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