Both car A and car B set out from their original locations

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