Question

41. Suppose a family drives at an average rate of 60 mi/h on the way to visit relatives and then at an average rate of 40 mi/h on the way back. The return trip takes 1 h longer than the trip there.

a. Let d be the distance in miles the family traveled to visit their relatives. How

many hours did it take to drive there?

b. In terms of d, how many hours did it take to make the return trip?

C. Write and solve an equation to determine the distance the family drove to see

their relatives. What was the average rate for the entire trip?

Answer 1

The answer is below

Explanation:

a) Let x represent the time taken to drive to see the relatives and let d be the distance travelled to go, hence:

60 mi/h = d/x

d = 60x

When returning, they still travelled a distance d, since the return trip takes 1 h longer than the trip there, therefore:

40 mi/h = d/(x+1)

d = 40(x + 1) = 40x + 40

Equating both equations:

60x = 40x + 40

60x - 40x = 40

20x = 40

x = 40/20

x = 2 h

The time taken to drive there = x = 2 hours

b) The time taken for return trip = x + 1 = 2 + 1 = 3 hours

c) The distance d = 60x = 60(2) = 120 miles

The total distance to and fro = 2d = 2(120) = 240 miles

The total time to and fro = 2 h + 3 h = 5 h

Average speed = total distance / total time = 240 miles / 5 h = 48 mi/h