Systems Equations

Question

A motorboat travels 180 kilometers in 6 hours going upstream. It travels 348 kilometers going downstream in the same amount of time. What is the rate of the boat in still water and what is the rate of the current?

Answer 1

d=rt

180=(r-w)6

30=r-w

348=(r+w)6

58=r+w

add the two equations

30+58= 2r

r = 88/2 =44 km/hr = boat speed, in still water

w = r-30= 14 km/hr = current speed

r=rate of speed of the boat in still water

w= speed of the water current

up stream speed = r-w = 30 km/hr

down stream speed = r+w = 58 km/hr

in 6 hours, the boat goes 6(30) = 180 km up stream

in 6 hours the boat goes 6(58) = 348 km down stream

Answer 2

Here is another way to approach the problem.

Find the average of the two distances (each takes the same time)

(180 + 348)/2= 528/2 = 264.

So with NO current, the boat's speed must be 264/6 = 44 km/hr. That's the answer to the 1st question.

Against the current the speed was 180/6 = 30 km/hr, so the current must have slowed the boat's progress by 44 - 30 = 14 km/hr.

Answer 3

180 = 6(b - c), b represents the boat, c represents the current

348 = 6(b + c)

Can you solve the system and answer?