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Math Help - is this right?

  1. #1
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    Question is this right?

    A cyclist and a jogger leave at the same time from the same spot and travel the same path. The cyclist rides 6 mph faster than the jogger runs. It takes the jogger 2 hrs to do the same distance as the cyclist does in 1 hr. find the speed of each.

    I drew my box and did the operation D=R * T and came up with

    6(R+1) = 2(R)
    coming up with a final answer of R = -2

    Can you tell me if I did this right? Thanks so much!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tpoma View Post
    A cyclist and a jogger leave at the same time from the same spot and travel the same path. The cyclist rides 6 mph faster than the jogger runs. It takes the jogger 2 hrs to do the same distance as the cyclist does in 1 hr. find the speed of each.

    I drew my box and did the operation D=R * T and came up with

    6(R+1) = 2(R)
    coming up with a final answer of R = -2

    Can you tell me if I did this right? Thanks so much!!
    obviously it is not right. first, you have one answer, they asked for the rate of each, which means you should have two answers. secondly, you have a negative answer. which makes no sense for this problem. were they cycling/jogging backwards?

    go with your set up again. let R be the rate of the cyclist. then R - 6 is the rate of the jogger. if the cyclist takes T hours to cover a distance D, then the jogger takes 2T hours to cover the same distance.

    now what equations would you set up?
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  3. #3
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    Hello, tpoma!

    Something's wrong . . . The speed should not be negative.


    A cyclist and a jogger leave at the same time from the same spot
    and travel the same path. The cyclist rides 6 mph faster than the jogger runs.
    It takes the jogger 2 hrs to do the same distance as the cyclist does in 1 hr.
    Find the speed of each.
    I'm not sure how you set up your "box" . . .


    Let x = speed of the runner.
    Then x+6 = speed of the cyclist.

    \begin{array}{|c||c|c|c|c|c} & R & \times & T & = & D \\ \hline \text{Runner} & x & & 2 & & 2x \\ \hline \text{cyclist} & x + 6 & & 1 & & x+6 \\ \hline \end{array}
    . . . . . . . . . . . . . . . . . . . \Uparrow
    . . . . . . . . . . .
    These distances are equal.


    There is our equation! . . . . 2x \:=\:x+6

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  4. #4
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    Ok I think I got it.........

    Is this right............

    2(R-6) = 1R
    2R -12=1R
    2R-1R=12
    R=12
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  5. #5
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    Red face

    So that's it that's the answer for both questions???? 2R=R+6 Thats all they wanted to know?? I don't have to solve it down anymore?

    If not thank you so much for all your help! I didn't think I was ever going to get this one! You people ROCK!!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tpoma View Post
    So that's it that's the answer for both questions???? 2R=R+6 Thats all they wanted to know?? I don't have to solve it down anymore?

    If not thank you so much for all your help! I didn't think I was ever going to get this one! You people ROCK!!
    again, you should have two answers: you need the cyclist's speed, and the jogger's speed. see Soroban's post
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  7. #7
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    Question

    Ok so one of the two answers is X = 6 still working on answer # 2
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tpoma View Post
    Ok so one of the two answers is X = 6 still working on answer # 2
    what is there to work on? going by Soroban's work, one rate is x, the other is x + 6. also, you have to say which is which
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  9. #9
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    cyclist speed is R+6=6+6=12

    joggers speed is 6 mph

    RIGHT
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  10. #10
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    Quote Originally Posted by tpoma View Post
    cyclist speed is R+6=6+6=12

    joggers speed is 6 mph

    RIGHT
    right
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