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Ray who lives in the countryside, caught a train for home ea

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Ray who lives in the countryside, caught a train for home ea [#permalink] New post 24 Sep 2009, 04:27
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Question Stats:

34% (05:01) correct 65% (03:29) wrong based on 84 sessions
Ray who lives in the countryside, caught a train for home earlier than usual yesterday.His wife normally drives to the station to meet him. But yesterday he set out on foot from the station to meet his wife on the way.He reached home 12 mins earlier than he would have done had he waited at the station for his wife. The car travels at uniform speed which is 5 times Ray's speed on foot. Ray reached home exactly at 6' o clock. At what time would he have reached home if his wife, forewarned of his plan, had met him at the station.

A. 5:48
B. 5:42
C. 5:00
D. 5:36
[Reveal] Spoiler: OA
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Re: Time speed 5 [#permalink] New post 24 Sep 2009, 06:49
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I got 5.36

take: a=distance covered by ray's walk
x=ray walk's speed
5x=car's speed
12 mins is the time needed by the car to cover twice a (car met ray in the middle)
2a=5x.12
a=30x -->ray walk for 30 mins

If car meet ray 30mins earlier in station then for a, car only need t=30x/5x=6 mins
Time saved is 30 mins minus 6 mins which is 24 mins
so instead of 6, he could arrive at 5.36

OA please
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Re: Time speed 5 [#permalink] New post 03 Oct 2009, 19:06
myellen wrote:
I got 5.36
12 mins is the time needed by the car to cover twice a (car met ray in the middle)


How did u deduce this ? It's not mentioned anywhere in the problem that the car met ray in the middle
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Re: Time speed 5 [#permalink] New post 16 Aug 2010, 05:05
Can someone please explain the solution? Bunuel?
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Re: Time speed 5 [#permalink] New post 16 Aug 2010, 08:35
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nonameee wrote:
Can someone please explain the solution? Bunuel?

Ray who lives in the countryside, caught a train for home earlier than usual yesterday. His wife normally drives to the station to meet him. But yesterday he set out on foot from the station to meet his wife on the way. He reached home 12 mins earlier than he would have done had he waited at the station for his wife. The car travels at uniform speed which is 5 times Ray's speed on foot. Ray reached home exactly at 6' o clock. At what time would he have reached home if his wife, forewarned of his plan, had met him at the station.

As they "reached home 12 mins earlier than he would have done had he waited at the station for his wife" means that usually they reach home at 6:12 (6:00+12min). Now if we knew how many minutes earlier did Ray arrive at the station than 6:12 minus that time would be the time he have reached home if his wife, forewarned of his plan, had met him at the station.

As they arrived 12 minutes earlier than usual, wife saved 12 minutes on round trip from home to station (home-station-home) --> 6 minutes in each direction (home-station) --> so wife met husband on the road 6 minutes earlier than the usual time of their meeting on the station.

As wife met husband 6 minutes earlier than the usual time of their meeting --> so distance usually covered by wife in 6 minutes was covered by Roy --> as rate of wife is 5 times that of Roy's then Roy spent 5*6=30 minutes covering that distance. So Roy arrived 30 minutes + 6 minutes = 36 minutes earlier than usually.

6:12 - 36 minutes = 5:36.

Answer: D.

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ray-who-lives-in-the-countryside-caught-a-train-for-home-ea-84295.html
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Hope it helps.
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Re: Time speed 5 [#permalink] New post 18 Aug 2010, 03:57
Bunuel, thanks a lot. I got it. Is it me, or is this question really difficult?
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Re: Time speed 5 [#permalink] New post 04 May 2014, 05:18
why did you add 6min in the end to make the total to 36min? Bunuel pls explain
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Re: Time speed 5 [#permalink] New post 04 May 2014, 07:59
rohan567 wrote:
why did you add 6min in the end to make the total to 36min? Bunuel pls explain


i will try to explain
he walked for 30 mins from station to where he met his wife , and the distance traveled by him in the 30 mins is traveled by his wife in 6 min
and as his wife took 12 mins (home - where husband was standing - home ) 12/2 =6 mins one way on an avg
so it will still take them 6 mins to reach home
therefore, 30 mins he traveled by foot + 6 mins he traveled in the car = 36 mins
6:12-0:36 = 5:36
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Re: Time speed 5 [#permalink] New post 08 Jun 2014, 09:55
Bunuel wrote:
nonameee wrote:
Can someone please explain the solution? Bunuel?

Ray who lives in the countryside, caught a train for home earlier than usual yesterday. His wife normally drives to the station to meet him. But yesterday he set out on foot from the station to meet his wife on the way. He reached home 12 mins earlier than he would have done had he waited at the station for his wife. The car travels at uniform speed which is 5 times Ray's speed on foot. Ray reached home exactly at 6' o clock. At what time would he have reached home if his wife, forewarned of his plan, had met him at the station.

As they "reached home 12 mins earlier than he would have done had he waited at the station for his wife" means that usually they reach home at 6:12 (6:00+12min). Now if we knew how many minutes earlier did Ray arrive at the station than 6:12 minus that time would be the time he have reached home if his wife, forewarned of his plan, had met him at the station.

As they arrived 12 minutes earlier than usual, wife saved 12 minutes on round trip from home to station (home-station-home) --> 6 minutes in each direction (home-station) --> so wife met husband on the road 6 minutes earlier than the usual time of their meeting on the station.

As wife met husband 6 minutes earlier than the usual time of their meeting --> so distance usually covered by wife in 6 minutes was covered by Roy --> as rate of wife is 5 times that of Roy's then Roy spent 5*6=30 minutes covering that distance. So Roy arrived 30 minutes + 6 minutes = 36 minutes earlier than usually.

6:12 - 36 minutes = 5:36.

Answer: D.


Hi Bunuel,
I still don't get why 6 mins are added to 30 mins (that Ray walked) ?

Let's assume the wife takes 1 hr to drive from home to station.
Normal Scenario:
Wife leaves home at 4:12 pm, reaches station at 5:12 pm (picks up Ray) and then reaches home at 6:12 pm.

Early arrival Scenario:
Wife leaves home at 4:12 pm, reaches mid-way point at 5:06 pm (Picks up Ray), and then reaches home at 6:00 pm. (6 mins saved on each trip. 12 mins total)

Ray has walked for 30 mins. So, Ray arrived at the station at 5:12 - 30 mins = 4:42 pm.

So,
Wife (now knowing that Ray is taking the earlier train)
leaves home at 3:42 pm, reaches station at 4:42 pm (Picks up Ray), reaches home at 5:42 pm.

Please explain.

Regards,
Nikhil.
Re: Time speed 5   [#permalink] 08 Jun 2014, 09:55
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