Solving a Distance, Rate, Time Problem Using a Linear Equation. Word Problem

Question

Kaitlin drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 10 hours. When Kaitlin drove home, there was no traffic and the trip only took 7 hours. If her average rate was 18 miles per hour faster on the trip home, how far away does Kaitlin live from the mountains?

Do not round.

Answer 1

Hi Savannah:

Here's the solution.

We need the standard equation: s = vt where s = distance traveled, v = average speed, t = time

Let Kaitlin's average speed on the way to the mountain be v => on the way back, average speed = v + 18

Let the distance to the mountains be s.

Plugging into the standard equation, on the way to the mountain s = 10 v . . . . . . . . . . . (1)

And on the way back s = 7 (v + 18) . . . . . . . . (2)

Substitute (1) into (2): 10v = 7(v + 18)

10v = 7v + 126

3v = 126

v = 42 mph

Substitute into (1): s = 10 x 42 = 420 miles

Check: 420 miles at 42 mph = 10 hours and 420 miles at 42 + 18 = 60 mph = 7 hours, as required.