Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph.Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph. If both travelers maintained their speed and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?
A 35 mins
B 45 mins
C 1 hour
D 1 hour 15 mins
E 1 hour 30 mins
Let's say Mary passes the station at 1:45pm.
Implication:
At 2pm -- 15 minutes later -- Paul is AT the gas station, whereas Mary has traveled for 15 minutes BEYOND the gas station.
Since Mary's rate = 50mph, the distance traveled by Mary in 1/4 of an hour = rt = (50)(1/4) = 12.5 miles.
Paul has to CATCH-UP by 12.5 miles.
The CATCH-UP rate is equal to the DIFFERENCE between the two rates.
Since Paul's rate = 60mph, while Mary's rate = 50mph, every hour Paul travels 10 more miles than Mary, with the result that every hour Paul catches up by 10 miles -- the DIFFERENCE between the two rates:
60 - 50 = 10mph.
Thus:
Time for Paul to catch-up = (catch-up distance)/(catch-up rate) = (12.5)/(10) = 1.25 hours = 1 hour, 15 minutes.
The correct answer is D.
