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Manager
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Mary on the highway [#permalink]
 01 May 2010, 09:31
Question Stats:
50% (02:21) correct
50% (02:41) wrong based on 2 sessions
Mary passed a certain gas station on a highway while traveling west at a constant
speed of 50 miles per hour. Then, 15 minutes later, Paul passed the same gas station
while traveling 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. 30 min
B. 45 min
C. 1 hr
D. 1 hr 15 min
E. 1 hr 30 min
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Current Student
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Re: Mary on the highway [#permalink]
01 May 2010, 10:54
When Paul is at the station, Mary is 50/4 miles ahead on the highway. (the distance she drove in 15 min)
every hour, paul drives 10miles more than Mary. how many hours will it takes him to drive 50/4 miles more? the answer is (50/4)/10 = 25/20=1h15min.
Answer D
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Senior Manager
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Re: Mary on the highway [#permalink]
01 May 2010, 11:28
Net distance traveled by Mary in 15 mins = 50*15/60
So relative distance between mary and Paul, when paul passed through station =50/4
Net time to meet mary is = relative distance/relative speed = (50/4)/(60-50) = 5/4 = 1+1/4 = 1 hr 15 min
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CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
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Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
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Re: Mary on the highway [#permalink]
01 May 2010, 15:47
Relative speed = 60-50 = 10miles/hour relative distance between them = distance traveled by Mary in 15 min = 50 * 15/60 50 * 15/60 = 10 t => t = 1 hour 15 mins
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Re: Mary on the highway
[#permalink]
01 May 2010, 15:47
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