Find Huong's average speed on the outbound trip.

Question

Huong drove to the ferry office and back.
The trip there took one hour and the trip
back took 3.3 hours. She averaged 46
km/h faster on the trip there than on the
return trip. 

Answer 1

Remember that distance = rate × time. We'll write two equations, one for the outbound trip and one for the return trip.

 

 

 

OUTBOUND

 

We're not told how far she traveled, so call the distance d. We're not told how fast she traveled, so call the rate r. She took 1 hour to travel there. So:

 

d = 1r

d = r

 

 

 

RETURN

 

We're not told how far she traveled, but she traveled the same distance as her trip there, so call the distance d. We're not told how fast she traveled, but she traveled 46 kph slower on the return trip, so call the rate r - 46. She took 3.3 hours to travel back. So:

 

d = 3.3(r - 46)

 

 

 

Now you have a system of two equations:

 

d = r

d = 3.3(r - 46)

 

Substitute the first equation into the second equation for d, then solve for r:

 

r = 3.3(r - 46)

r = 3.3r - 151.8

-2.3r = -151.8

r = 66

 

Huong averaged 66 km/h on the outbound trip.