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Math Help - trip calculation

  1. #1
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    Help!!!I dun understand the question

    Marcus and Bryan decided to visit one another and set off at the same time to each otherís house. The distance between their houses is 7 km. When they met on the road joining their houses, they forgot that they wanted to see each other and continued walking. After they met, it took Bryan nine times as long to reach Marcusís house as it took Marcus to reach Bryanís house. What is the distance between Bryanís house and the point where he met Marcus?
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  2. #2
    Junior Member hoeltgman's Avatar
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    You are asked to find out where the two met. (you know the total distance and how far they are away from each others house. That's all you need) And finally how far Bryan is away from his home.
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  3. #3
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    Dun understand

    but i dunnoe the speed
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  4. #4
    Junior Member hoeltgman's Avatar
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    Ok look at this:

    The distance between their houses is 7 km.
    After they met, it took Bryan nine times as long to reach Marcusís house as it took Marcus to reach Bryanís house.

    That's all you need to find out the position.

    \dfrac{10}{10} = 7000m

    We know that they have met and stand at the same place. In that case:
    distance for Bryan to Marcus' House :  \dfrac{9}{10} \cdot 7000 = 6300m
    distance for Marcus to Bryan's House:  \dfrac{1}{10} \cdot 7000 = 700m

    Well now you know the positions!

    What is the distance between Bryanís house and the point where he met Marcus?

    If Marcus has to go 700m to reach Bryan's House. Well then the solution is simply 700m.
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  5. #5
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    thx:)

    oo i got this solution, but thought how can it be so easy
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