Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 years ago

A motorboat traveled 48 mi down river in 2 hrs. The return trip takes 3 hrs.?

What was the rate of the boat in still water? What was the rate of the current?

a. Boat in still water: 4 mi/hr, Current: 10 mi/hr

b. Boat in still water: 10 mi/hr, Current: 4 mi/hr

c. Boat in still water: 4 mi/hr, Current: 20 mi/hr

d. Boat in still water: 20 mi/hr, Current: 4 mi/hr

3 Answers

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  • Robert
    Lv 6
    8 years ago
    Favorite Answer

    2 = 48/(S+C)

    3 = 48/(S - C)

    2S + 2C = 48 ------>> 6S + 6C = 144

    3S - 3C = 48 ------>> 6S - 6C = 96

    12S = 240

    S = 20

    2S + 2C = 48 ------>> S + C = 24, C = 4

    d. Boat in still water: 20 mi/hr, Current: 4 mi/hr

  • 8 years ago

    Assume speed of boat = x mph and speed of current= y mph

    Going upstream:

    48/(x - y) = 3

    Going downstream:

    48/(x + y) = 2

    2x = 40 => x = 20 mph

    So only option D works.

  • 3 years ago

    formulation of velocity: v=d/t, the place V is velocity, D is distance and T is time. doing a sprint rearranging we get that t=d/v In equation decrease than v is velocity of the boat and r is the fee of the flow 40 8/(v+r)=3 40 8/(v-r)=12 fixing that equation we get that r=6

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