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Math Help - Word Problem #5

  1. #1
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    Word Problem #5

    Chrstobal decided to make a 28km canoe trip up and 28km trip down the Grand River. The speed of the current is3km. One way, Christobal travelled with the current and the other way against the current. During his trip, Christobal stopped half an hour for lunch. If Christobal's total trip lasted 6 hours, how fast did he travel?
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  2. #2
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    Quote Originally Posted by Chrissy
    Chrstobal decided to make a 28km canoe trip up and 28km trip down the Grand River. The speed of the current is3km. One way, Christobal travelled with the current and the other way against the current. During his trip, Christobal stopped half an hour for lunch. If Christobal's total trip lasted 6 hours, how fast did he travel?
    Hello,

    nearly the same problem was answered by Capt.Black ("Another word problem" by sugar_babee). So I'll give you only the equation which you can solve easily:

    The time of canoeing adds up to 5.5 h. So you get:
    {28 \over {v+3}}+{28 \over{v-3}}=5.5

    The lcd = (v+3)(v-3). Multiply by lcd and rearrange the terms, then you have:

    5.5 \cdot v^2-56v-49.5=0

    You'll get 2 answers. You have to check which one is plausible with your problem.

    Greetings

    EB
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