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27 June, 15:19

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?

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  1. 27 June, 15:38
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    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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