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18 June, 03:02

Two cyclists leave towns 120 km apart at the same time and travel toward each other. One cyclist travels 10 km/h slower than the other. If they meet in 2 hours, what is the rate of each cyclist?

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Answers (2)
  1. 18 June, 03:04
    0
    Exp35 km/h, 25 km/hlanation:

    Let the speed of one cyclist is v and the speed of another cyclist is v - 10.

    Distance traveled by the first cyclist in 2 hours = v x 2

    Distance traveled by the second cyclist in 2 hours = (v - 10) x 2

    They meet after 2 hours.

    So the distance traveled by the first and the second cyclist is equal to 120 km.

    2 v + 2 (v - 10) = 120

    2v + 2v - 20 = 120

    4v = 140

    v = 35 km/h

    The speed of first cyclist is 35 km/h and the speed of another cyclist is 35 - 10 = 25 km/h.
  2. 18 June, 03:09
    0
    35 km/hr and 25 km/hr

    Explanation:

    Total distance is D.

    Let speed of one cyclist be X km/hr.

    Speed of second cyclist = (X-10) km/hr

    Now,

    Distance = Speed * Time

    Total distance D = Distance covered by slower cyclist + Distance covered by faster cyclist.

    D = 2 * X + 2 (X-10)

    120 = 2X + 2X - 20

    120 = 4X - 20

    140 = 4X

    X = 35 km/hr

    Therefore, Speed of faster cyclist = 35 km/hr

    Speed of slower cyclist = 35-10

    = 25km/hr
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