Question

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 same 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?

Answer 1

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.