Ask Question
21 December, 23:43

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?

+3
Answers (1)
  1. 22 December, 03:05
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers