Question

Mrs.Watt traveled between two cities that were 750 km apart. On the return trip, she increased her average rate of travel by 20 km/hr and made the trip in 10 hours less time. Find her rate of travel in each direction.

Answer 1

Let s = his speed on the outbound trip

Then s + 20 = his speed on the return trip

Let t = his travel time on the outbound trip

Then t - 10 = his travel time for the return trip

Given: d = 750

Since distance = speed x time, for the two trips we have

750 = s*t

750 = (s+20)(t-10)

Solve for t in the 1st equation, substitute in the 2nd:

750 = (s+20)(750/s-10)

Solve for s:

750 = 750 - 10s + 15000/s - 200

Simplify:

s^2 + 20s - 1500 = 0

Factor:

(s+50)(s-30) = 0

Take the positive solution, s = 30

So the rates are 30 kph and 50 kph