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11 March, 06:39

Two automobiles leave the same city simultaneously and both head towards another city. The speed of one is 10 km/hour greater than the speed of the other, and this is why the first automobile arrives at the destination 1 hour before the other. Find the speed of both automobiles knowing that the distance between the two cities is 560 km.

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  1. 11 March, 08:04
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    speed for the slower automobile = 70

    speed for the faster automobile = 80

    Step-by-step explanation:

    Speed of slower automobile = s and time taken by the slower automobile = t

    Speed for faster automobile = s + 10 and time taken by the faster automobile = t - 1

    D = s*t

    For the slower automobile

    560 = st

    For the faster automobile

    560 = (s + 10) (t-1)

    560 = st - s + 10t - 10

    Remember st = 560

    560 = 560 - s + 10 (560/s) - 10

    560-560 + 10 = - s + 5600/s

    10 = - s + 5600/s

    10s = - s² + 5600

    s² + 10s - 5600 = 0

    s = (-b ± (√b²-4ac)) / 2a

    s = (-10±√ (10² - 4 (1) (-5600))) / 2

    s = (-10 ±√22500) / 2

    s = (-10+150) / 2 or (-10 - 150) / 2

    s=70 or - 80

    therefore speed for the slower automobile = 70

    and speed for the faster automobile = 80
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