Need Help Solving a Distance Problem. D=RT? Distnace Rate Time?

Question

May drove to an appointment in the morning and traveled at 50mph. On the return trip her average speed was 40mph and the trip took 1/5 of an hour longer. How far did she travel to the appointment?

I have been playing with it and have no idea how to set it up. Thank you in advance for your help!

Answer 1

Let T be the time to get to the appointment (in hours):
Let T + 1/5 be the time to get back from the appointment:

Distance to the appointment:
D = 50T

Distance back from the appointment:
D = 40(T + 1/5)
D = 40T + 8

Since the distance is the same both ways, equate them:
50T = 40T + 8
10T = 8
T = 8/10
T = 4/5

Using either equation, now solve for the distance D:
D = 50T
D = 50(4/5)
D = 40 miles

In summary:
May took 4/5 hour at 50 mph to drive to the appointment (40 miles).
On the return she took 1 hour at 40 mph to go those same 40 miles.

Answer 2

Distance = rate x time

Since the distance is the same, you can set it up as an equality:

50 t = 40 x (t + 0.20) (1/5 of an hour is 0.20 of an hour)

50t = 40t + 8

10t= 8

t = 0.8 hour

Substitute for t in the distance equation, and you get

distance = 50 x 0.8

distance = 40 miles

Answer 3

d = 50 t (outbound trip)
d = 40 ( t + .2) (return trip - .2 hours longer)
d = 40 t + 8(multiply out last statement.)

40 t + 8 = = 50 t
8 = 50 t - 40 t
8 = 10 t
t = .8 hours. So, she took .8 hours on the outbound trip.
substitute back into first statement.
d = 50 (.8)
d = 40 km

Answer 4

50t=40(t+1/5) because the distance is the same
50t=40t+8
t=.8hours
so she drove 40miles