Bunuel
Car A and Car B are each traveling along the same stretch of highway in the same direction. Car A leaves at 10:00am traveling at a constant rate of 40 miles per hour and Car B leaves at 10:30am traveling at a constant rate of 50 miles per hour. At what time will Car B be 40 miles ahead of Car A?
A. 3:30pm
B. 4:00pm
C. 4:30pm
D. 5:00pm
E. 5:30pm
When Car B Starts at 10:30, Car A has already covered 20 Miles(Speed is 40 Miles per hour).
=> The gap between the two is 20 miles.
Thus, Car A and Car B will meet after time = Gap/Difference in speeds ( Difference is taken as both are travelling in the same direction)
=> Time = 20/50-40=20/10=2 Hours.
Thus, both the cars will meet after 2 hours. Now , in order to make a gap of another 40 miles, another period of time required = 40/50-10=4 hours.
Thus , a total of 2+4=6 hours is required for making Car B 40 miles ahead of Car A.
Thus, 10:30 a.m. + 6 hours = 4:30 p.m.
-- Hit Kudos if you get the answer