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March 28, 2017

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mr kumar lives in the eastern part of spore . he visits his aged parents , who lives 36 km away , every weekend he finds that is he increases the average speed of his vehicle by 12km/h, he could save 9 minutes of his travelling time. find the speed at which he travels before the increase in speed

  • MAth grade 8 - ,

    speed before increase : x km/h
    time for trip with old speed = 36/x hrs

    speed after increase : x+12 km/h
    time for trip after increase = 36/(x+12)

    36/x - 36/(x+12) = 9/60 = 3/20
    divide by 3
    12/x - 12/(x+12) = 1/20
    multiply be 20x(x+12)
    240(x+12) - 240x = x(x+12)
    240x + 2880 - 240x = x^2 + 12x
    x^2 + 12x - 2880 = 0
    x^2 + 12 + 36 = 2880 + 36
    (x+6)^2 = 2916
    x+6 = ± √2916 = ±54
    x = -6 ± 54
    x = 48 or a negative

    So he originally travelled at 48 km/h

    check:
    first time = 36/48= .75 hrs = 45 minutes
    second time = 36/60 = .6 hrs = 36 minutes,
    sure enough, the difference he gains is 9 minutes.

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