Problem Solving

rsarashi wrote:Mary passed a certain gas station on a highway while travelling west at a constant speed of 50 miles per hour. Then, 15 minutes later, Paul passed the same gas station while travelling west at a constant speed of 60 miles per hour. 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) 30min

B) 45min

C) 1hr

D) 1hr 15min

E) 1hr 30min

OA
D

Mary passes gas station and travels at 50 mph for 15 minutes
So, Mary drove for 1/4 hours
Distance traveled = (time)(speed)
= (1/4)(50)
= 50/4
= 12.5 miles

Paul passes the same gas station and travels at 60 mph
IMPORTANT: At this point, Paul is 12.5 miles behind Mary.
However, since Paul is traveling 10 mph FASTER than Mary, the gap between them SHRINKS at a rate of 10 mph.
So, at this point, we need to determine the time it takes for the 12.5 mile gap to shrink to zero.
Time = (distance)/(speed)
= 12.5/10
= 1.25 hours
= 1 hour 15 mins
= D

Cheers,
Brent

Hi rsarashi,

Your prompt has a 'typo' in it (it states that Mary drove for 1 minute; it should state that she drove for '15 minutes').

GMAT assassins aren't born, they're made,
Rich

rsarashi wrote: ↑

Sat Sep 02, 2017 11:56 am

Mary passed a certain gas station on a highway while travelling west at a constant speed of 50 miles per hour. Then, 15 minutes later, Paul passed the same gas station while travelling west at a constant speed of 60 miles per hour. 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) 30min

B) 45min

C) 1hr

D) 1hr 15min

E) 1hr 30min

OA
D

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