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Math Help - Difficult word problem...

  1. #1
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    Difficult word problem...

    An army begins marching in a single file, straight line that is 40 miles long. They march for 40 miles. In the time that it takes the army to march 40 miles, the soldier who is last in line rides a horse to the front of the line to deliver a message; then rides his horse back to his position at the back of the line. He returns to his position in line at the same time the army completes its 40th mile. How many total miles does the horse run?
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  2. #2
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    80, because the horse rides to the front once and to the back once then he continues on foot.
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  3. #3
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    Quote Originally Posted by Emergency View Post
    80, because the horse rides to the front once and to the back once then he continues on foot.
    Im sorry but this most sertainly is not correct.
    It said "he returns to his position in the line". He does not return to the origin.
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  4. #4
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    Code:
    A:@a....y....[1]...40-y...[3]
     
    B:@b................40+y...............[2]
    A=back of army line (speed = a), B=horse position (speed = b)

    A travels distance y, to position [1]
    B travels 40 miles further (to end of army line) to position [2]
    A arrives at [1] at same time as B arrives at [2]

    A continues such that it travels total of 40, to position [3]
    B comes back, to position [3] : both get there at same time

    You now have enough info to get b = a[1 + SQRT(2)]
    From that, you can calculate y
    Distance by horse = x + 2y

    That's all I'll give you; don't wanna spoil your fun of getting b, then y
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  5. #5
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    Quote Originally Posted by Wilmer View Post
    Code:
    A:@a....y....[1]...40-y...[3]
     
    B:@b................40+y...............[2]
    A=back of army line (speed = a), B=horse position (speed = b)

    A travels distance y, to position [1]
    B travels 40 miles further (to end of army line) to position [2]
    A arrives at [1] at same time as B arrives at [2]

    A continues such that it travels total of 40, to position [3]
    B comes back, to position [3] : both get there at same time

    You now have enough info to get b = a[1 + SQRT(2)]
    From that, you can calculate y
    Distance by horse = x + 2y

    That's all I'll give you; don't wanna spoil your fun of getting b, then y
    I was able to come up with basically the same set up too, except I have the Distance by horse = 40 + 2y. I could be wrong. I still don't know what to do from here. What is "a[1 + SQRT(2)]"??? I am able to follow your logic. I just don't know how to set up an equation with the given information to solve for the variables.
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  6. #6
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    Quote Originally Posted by bearej50 View Post
    I was able to come up with basically the same set up too, except I have the Distance by horse = 40 + 2y
    Yes, that's what I meant...was using x as army line length...
    x = 40 ; from [1][2] :
    (x + 2y) / b = x / a ; leads to x = 2ay / (b-a)
    From [3]:
    (x - y) / a = y / b ; leads to x = (ay + by) / b
    So:
    2ay / (b-a) = (ay + by) / b
    The y's cancel out and we get: b^2 - 2ab - a^2 = 0
    Using quadratic: b = a +- aSQRT(2)
    "-" impossible, so b = a[1 + SQRT(2)]

    Substitute back and you should be ok.
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