Question

A canoeist travels 30 miles downstream in three hours. Against the current the return trip took fifteen hours. Find the rate of the canoeist in calm water and the rate of the current. ​

Answer 1

Answer: the rate of the canoeist in calm water is 6 mph and the rate of the current is 4 mph

Step-by-step explanation:

Let x represent the rate of the canoeist in calm water.

Let y represent the rate of the current.

The canoeist travels 30 miles downstream in three hours.. Assuming that he travelled in the direction of the current, his total speed would be x + y mph

Distance = speed × time

Distance covered downstream is

30 = 3(x + y)

10 = x + y - - - - - - - - - - - -1

Against the current the return trip took fifteen hours..

His total speed would be x - y mph

Distance = speed × time

Distance covered on return trip is

30 = 15(x - y)

2 = x - y - - - - - - - - - - - -2

Adding equation 1 and equation 2, it becomes

12 = 2x

x = 12/2 = 6 mph

Substituting x = 6 into equation 1, it becomes

10 = 6 + y

y = 10 - 6 = 4 mph

Answer 2

The rate of the canoeist in calm water is 6 miles per hour while the rate of the current is 4 miles per hour

Speed is the ratio of the distance travelled to time taken. It is given by:

Speed = distance / time

Let a represent the speed of the canoeist and b represent the speed of the current

A canoeist travels 30 miles downstream in three hours, hence:

(a + b)3 = 30

a + b = 10    (1)

Against the current the return trip took fifteen hours, hence:

(a - b)15 = 30

a - b = 2    (2)

Solving equation 1 and 2 simultaneously gives:

a = 6 miles per hour and b = 4 miles per hour

The rate of the canoeist in calm water is 6 miles per hour while the rate of the current is 4 miles per hour

Find out more at: https://brainly.com/question/22610586