MATH PROBLEM DOWN THERE! :)?

A motorboat travels 306 km in 6 hours going upstream and 873 km in 9 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?

Rate of the boat in still water: __km/h

Rate of the current: __km/h

5 Answers

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  • 9 years ago
    Favorite Answer

    Let b = boat's rate in still water and c = current's rate.

    6(b - c) = 306

    9(b + c) = 873

    solve the system of equations:

    6(b - c) = 306

    b - c = 51

    b = 51 + c

    9(51 + c + c) = 873

    51 + c + c = 97

    2c = 46

    c = 23

    b = 51 + 23 = 74

  • TBT
    Lv 7
    9 years ago

    Rate of the boat in still water: u km/h

    Rate of the current: v km/h

    306 / (u-v) = 6

    51= u - v ---- (1)

    873 / (u +v) = 9

    97 = u + v ---- (2)

    (1) +(2)

    148 = 2 u

    u = 74 km/h --answer

    51 = 74 - v

    v = 23 km/h ---answer

    Source(s): my brain (Prof TBT)
  • 9 years ago

    Let the speed of boat in still waters be x km per hour

    And rate of current be y km /hr

    UPSTREAM

    306/6 = 51 = x -y (A)

    Downstream

    873/9 = 97 = x+y (B)

    A +B = 2x = 148

    x = 74 km/ hour

    y = 23 km/hr ANSWER

  • mayi o
    Lv 4
    9 years ago

    going with the current: v + vc

    going against the current: v-vc

    going through still water: v

    306/6 = 51 km/hr

    873/9 = 97 km/hr

    v+vc = 97

    v-vc = 51

    2v = 148

    v = 74 km/hr

    rate of current, vc = 23 km/hr

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  • Mike G
    Lv 7
    9 years ago

    306 = 6(V-C) so 153 = 3V-3C

    873 = 9(V+C) so 291 = 3V+3C subtract equations

    138 = 6C

    C = 23 km/h = current

    V = 74 km/h = boat

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