aimeehittinger wrote:

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 drivers maintained their speeds 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. 30 min

B. 45 min

C. 1 hr

D. 1 hr 15 min

E. 1 hr 30 min

We can classify this problem as a “catch-up” rate problem, for which we use the formula:

distance of Mary = distance of Paul

We are given that Mary passed a certain gas station traveling at a rate of 50 mph and that 15 minutes later Paul passed the same gas station traveling at a rate of 60 mph.

Since Mary passed the gas station 15 minutes before Paul, we can let Paul’s time = t hours and Mary’s time = t + 1/4 hours.

Since distance = rate x time, we can calculate each distance in terms of t.

Mary’s distance = 50(t + 1/4) = 50t + 50/4 = 50t + 25/2

Paul’s distance = 60t

We can equate the two distances and determine t.

50t + 25/2 = 60t

25/2 = 10t

t = (25/2)/10 = 25/20 = 5/4 = 1¼ hours = 1 hour and 15 minutes

Answer: D

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