Question 258035: Please setup the equation and solve the following word problem:
Bob Rodriguez, the Albany, NY sales representative for Advanced Drafting Software, Inc. must attend the regional sales meeting in New York City. Unfortunately, his car is in the shop so he must make other travel arrangements to reach his destination. One option is to take a train that departs at 8 am. alternatively, he could ride to the meeting with another sales rep. Bob knows that the train travels at 80 miles per hour, but has a 30-min. layover halfway through the 156-mile trip. He also knows that his friend cannot depart before 8 am and will refuse to drive any faster than 55 mph. If both methods of transportation leave at the same time (8am) will the train or the care allow him to reach the meeting sooner?
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! They have tried to confuse you with all the verbage. Here is a table base on the given information, but we must assume distances are equal.
TRANSPORT . . . . . . rate . . . . . . time . . . . . . distance
CAR . . . . . . . . . .55 . . . . . . . .T . . . . . . . 156
TRAIN . . . . . . . . .80 . . . . . . . .T . . . . . . . .156
Tc = car time ~ 2.8363 hours
Tt = train time ~ 1.95 assuming it doesn't stop.
However the train stops for 1/2 hr. So add that to 1.93 to get 2.45
It is easy to see that Tt < Tc by about 23 minutes
Go with the train.
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