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A motorist averages 30 kph on ordinary loads and 12 kph on roads under repair. His average speed for a distance of 50 km is 24 km/h. What length of the road is under repair?

  • Math -


    time=50/24 hrs

    time rough+timesmooth= 50/24
    sum of distances equals 50km or
    timerough(30 -12)=50(1-1/2)

    time rough= 25/18 hrs
    distance rough= 25/18 * 12=50/3 km

    check all that.

  • Math -

    Let the distance on good road be x km
    then distance on rough road is 50-x km

    time on good road = x/30
    time on rough road = (50-x)/12

    total time = x/30 + (50-x)/12
    = (2x + 250 - 5x)/60
    = (250 - 3x)/60

    but total time = 50/24

    so (250-3x)/60 = 50/24
    3000 = 6000 - 72x
    72x = 3000
    x = 3000/72 = 125/3 or 41 .67 km

    so 42.67 km were good road and
    50 - 41.7 or 8.3 km were under repair.

    check: total time = 50/24 = 2.083 hrs

    time on good road = 41.67/30 = 1.389
    time on rought road = 8.3/12 = .692
    total time = 2.081 , close enough allowing for round-off
    all checks out

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