What would the system of equations be for the following question?

Posted on January 24, 2010 by admin

A riverboat travels downstream 8 km in 1 hour and returns the same distance in 1.5 hours.
Find the speed of the riverboat in still water and the speed of the river’s current; using a system of equations.

Please help!

Total Distance = 16Km
Total Time = 2.5Hr
Speed in still water = 16/2.5 = 6.4 Km/hr

‘Excess distance’ downstream = 8 – 6.4 = 1.6
Therefore current = 1.6Km/hr

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3 Responses to What would the system of equations be for the following question?

  1. GrahamH says:
    January 24, 2010 at 11:28 am

    Total Distance = 16Km
    Total Time = 2.5Hr
    Speed in still water = 16/2.5 = 6.4 Km/hr

    ‘Excess distance’ downstream = 8 – 6.4 = 1.6
    Therefore current = 1.6Km/hr
    References :

  2. peabody says:
    January 24, 2010 at 11:34 am

    x = riverboat speed and y = current

    Going downstream: d = vt
    8 = (x+y) 1
    8 = x + y (1)

    Going upstream: 8 = (x -y) 1.5
    16/3 = x -y (2)

    (1) + (2) gives 2x = 8 + 16/3 = 24/3 + 16/3 = 40/3

    x = 20/3

    Sub. x = 20/3 into (1) 20/3 + y = 8
    y = 8 – 20/3 = 24/3 – 20/3 = 4/3

    Therefore riverboats speed is 6 2/3 km/h and the current speed is 1 1/3 km/h.
    References :

  3. fuzzy says:
    January 24, 2010 at 12:16 pm

    Let B be speed of boat in still water and let R be speed of the river
    Speed of the boat going downstream= B+ R (river is pushing the boat)
    Speed of boat going upstream=B – R (river is working against the boat)

    speed= distance/time
    Speed of boat going downstream= 8km/1hr or 8 kph
    Speed of boat going upstream=8km/1.5hrs or 5.3 kph

    B+R=8kph
    B – R=5.3 kph

    Solve this system (add the two equations) and we get B= 6.6 kph and R= 1.3 kph
    References :