Question

Ted is a long distance driver. It took him 30 min longer to drive 275km on the Trans-Canada highway west of Swift Current, Saskatchewan, than it took him to drive 300km east of Swift Current. He averaged 10km/h less while travelling west of Swift Current due to more severe snow conditions. What was Ted's average speed for each part of the trip?

Answer 1

Speed traveling west = x - 10 km /hr

275 / ( x-10) - 300 / x = 1/2

275 *2x - 300 (x-10)*2 = ( x-10) *x

550x - 600x +6000 = x^2 -10x

x^2 +40x - 6000 = 0

( x -60) ( x+100) = 0

x = 60

ANSWER East = 60 km/ hr West = 50 km /hr

CHECK

300 /60 = 5 hours

275 /50 = 5.5 hours

Answer 2

The average speed on  east of Swift Current was 60km/h while it was 50km/h on the west of Swift Current

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

Speed = distance / time

Let t represent the time taken to drive on east of Swift Current.

Therefore for the west of Swift Current, time = t + 30 min = t + 0.5 hour

Let x be the average speed on  east of Swift Current.

Therefore for the west of Swift Current, average speed = x - 10

For West of swift current:

x - 10 = 275/(t + 0.5)

x = [275/(t + 0.5)] + 10   (1)

For east of swift current:

x  = 300/t     (2)

300/t = [275/(t + 0.5)] + 10

300t + 150 = 275t + 10t² + 5t

10t² - 20t - 150 = 0

t = 5 hours

x = 300 / t = 300/5 = 60 km/h

The average speed on  east of Swift Current was 60km/h while it was 50km/h on the west of Swift Current

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