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26 September, 14:33

The vagabonds were 200 miles apart at 2 p. M and were headed toward each other if they met at 6 p. M and one was traveling 10 miles per hour faster than the other what was the speed of each vagabond

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  1. 26 September, 15:23
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    First vagabond: 30 miles per hour

    Second vagabond: 20 miles per hour

    Step-by-step explanation:

    Let's call the speed of the first vagabond A, and the speed of the second vagabond B.

    They were heading toward each other, so the relative speed is the sum of both speeds: A+B

    The total number of hours spent until they meet is 6 - 2 = 4 hours, so we have the first equation:

    (A+B) * 4 = 200

    A + B = 50

    One vagabond was traveling 10 miles faster than the other, so we have the second equation:

    A = B + 10

    Using the value of A from the second equation in the first one, we have:

    B + 10 + B = 50

    2B = 40

    B = 20 miles per hour

    Now, from the second equation, we find the value of A:

    A = 20 + 10 = 30 miles per hour
  2. 26 September, 15:42
    0
    The answer is 200

    Step-by-step explanation:

    To solve the question given, let us recall the following,

    Let r4=200 miles apart

    r=50 mph combined speed

    Where

    2f+10=50

    2f=40

    f=20

    f+10=30

    or we can express it in another way,

    which is

    f x4 + (f+10) x 4=200

    f x4+f x4+40=200

    8f+40=200

    Then

    8f is = 160

    f=20

    f+10=30

    Therefore,

    4x 20+4 x 30=200

    80+120=200

    200=200
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