SOLUTION: Jane drove to an appointment averaging 60 mph. On the return trip, her average speed was 28 mph due to traffic, and she took 24 minutes longer. How far did she travel to the appoin
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Question 492943: Jane drove to an appointment averaging 60 mph. On the return trip, her average speed was 28 mph due to traffic, and she took 24 minutes longer. How far did she travel to the appointment? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Distance(d) equals Rate (r) times Time (t) or d=rt; r=d/t and t=d/r
Let d=distance to her appointment
And t=time required to travel to her appointment
Then d=60t
On the return trip:
d=28(t+24/60)=28(t+2/5) (Lets deal in hours for the time being )
So our equation to solve is:
60t=28(t+2/5)
60t=28t+56/5
32t=56/5
160t=56
t=56/160=7/20 hr or 21 min
d=60t=60*7/20 =21 mi
CK
at 28 mph on the return trip, it would take her
21/28 = 3/4 hr or 45 min
45-21=24 min longer
