swerve wrote:Both car A and car B set out from their original locations at exactly the same time and on exactly the same route. Car A drives from Morse to Houston at an average speed of 65 miles per hour. Car B drives from Houston to Morse at 50 miles per hour, and then immediately returns to Houston at the same speed and on the same route. If car B arrives in Houston 2 hours after car A, how many hours did it take car A to make its trip?
A. 0.50
B. 1.00
C. 1.25
D. 1.33
E. 2.00
We can let the distance between Morse and Houston = d miles. So the time it takes car A to drive from Morse to Houston is d/65. Since car B arrives in Houston 2 hours after car A and it also drives double the distance, we can create the following equation for the driving time:
2d/50 = d/65 + 2
Multiplying both sides by 650, we have:
26d = 10d + 1300
16d = 1300
d = 81.25
Therefore, it takes car A 81.25/65 = 1.25 hours to make the trip from Morse to Houston.
Alternate Solution:
Let the time car A takes to drive from Morse to Houston be t. Since car A drives at 65 mph, the distance between Morse and Houston, in terms of t, is 65t.
We are given that it takes t + 2 hours for car B to drive twice the distance between Morse and Houston; thus:
65t = [50(t + 2)]/2
130t = 50t + 100
80t = 100
t = 100/80 = 5/4 = 1.25 hours
Answer: C
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