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

Relevance

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

• 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.

• 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