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# Time speed 5

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Time speed 5 [#permalink]  24 Sep 2009, 05:27
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Q.
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.

1. 5:48
2. 5:42
3. 5:00
4. 5:36
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Re: Time speed 5 [#permalink]  24 Sep 2009, 07:49
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

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Re: Time speed 5 [#permalink]  24 Sep 2009, 10:56
I do agree with myellen's explanation.. Gr8 going.
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Re: Time speed 5 [#permalink]  25 Sep 2009, 13:51
OA

5.36 is the correct option

myellen wrote:
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

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Re: Time speed 5 [#permalink]  26 Sep 2009, 13:35
I too agree with myellen's explanation
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Re: Time speed 5 [#permalink]  03 Oct 2009, 20:06
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]  16 Aug 2010, 06:05
Can someone please explain the solution? Bunuel?
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Re: Time speed 5 [#permalink]  16 Aug 2010, 09:35
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.

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Re: Time speed 5 [#permalink]  18 Aug 2010, 04:57
Bunuel, thanks a lot. I got it. Is it me, or is this question really difficult?
Re: Time speed 5   [#permalink] 18 Aug 2010, 04:57
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