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Mary on the highway

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Manager
Joined: 09 Apr 2010
Posts: 79
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Mary on the highway [#permalink]  01 May 2010, 08:31
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Difficulty:

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Question Stats:

67% (02:29) correct 33% (02:41) wrong based on 6 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
Current Student
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Joined: 20 Jul 2009
Posts: 191
Location: Africa/Europe
Schools: Kellogg; Ross (); Tuck
Followers: 2

Kudos [?]: 32 [0], given: 6

Re: Mary on the highway [#permalink]  01 May 2010, 09:54
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.

Senior Manager
Joined: 24 Jul 2009
Posts: 297
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Kudos [?]: 99 [0], given: 0

Re: Mary on the highway [#permalink]  01 May 2010, 10:28
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
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2797
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 209

Kudos [?]: 1346 [0], given: 235

Re: Mary on the highway [#permalink]  01 May 2010, 14:47
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, 14:47
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