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A man arrives at a railway station 90 minutes before the tim

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A man arrives at a railway station 90 minutes before the tim  [#permalink]

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New post Updated on: 31 Aug 2010, 01:43
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A man arrives at a railway station 90 minutes before the time at which he had ordered his carriage to meet him. He walks at the rate of 4 kmph and meets the carriage 8km from his house and reaches 1 hour earlier than usual time. What is the speed of the carriage and distance from the station to his house?

A. 8kmph,12Km
B. 12kmph,16km
C. 4kmph,12km
D. None of these

Originally posted by RaviChandra on 30 Aug 2010, 05:17.
Last edited by RaviChandra on 31 Aug 2010, 01:43, edited 1 time in total.
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Re: Interesting Time Distance Question  [#permalink]

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New post 30 Aug 2010, 07:54
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RaviChandra wrote:
A man arrives at a railway station 90 before the time at which he had ordered his carriage to meet him. He walks at the rate of 4kmph and meets the carriage 8Km from his house and reaches 1 hour earlier than usual time. whats the speed of the carriage and distance from the station to his house.

1)8kmph,12Km
2)12kmph,16km
3)4kmph,12km
4)None of these


Not a GMAT question.

S----M--------H

As they arrived 60 minutes earlier than usual time then carriage saved 60 mins on the round trip Home-Station-Home. So carriage saved 60/2=30 mins in one way Home-Meeting point.

Suppose usual arrival time of that man at the station was 16:00 (this also would be usual meeting time of the man and the carriage), he arrived 90 minutes earlier than that so he arrived at 14:30.

Now, as carriage saved 30 minutes on the way from home to station, then carriage met the man 30 minutes earlier than their usual time of meeting, which is 16:00, so they met at 15:30. Which means that the man walked for 1 hour (he arrived at 14:30 and met the carriage at 15:30) and hence covered 1 hour*4 km/hour=4 km, so the distance between the station and the home is 4+8=12 km.

As for the speed of the carriage: this 4 km (SM) would be covered by carriage in 30 minutes so the speed of the carriage is 8 km/hour.

Answer: A.

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Re: Interesting Time Distance Question  [#permalink]

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New post 30 Aug 2010, 07:32
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RaviChandra wrote:
A man arrives at a railway station 90before the time at which he had ordered his carriage to meet him. He walks at the rate of 4kmph and meets the carriage 8Km from his house and reaches 1 hour earlier than usual time. whats the speed of the carriage and distance from the station to his house.

1)8kmph,12Km
2)12kmph,16km
3)4kmph,12km
4)None of these


I m assuming "90before the time" is 90 min before the time. I must say this is very poorly written question.

Home -------8------meet----d---station

According to the given condition he started moving towards the home and after traveling d km he met the carriage.
Time of travel = 90 min - 60 min = .5 hour
Speed = 4 km/h
Distance traveled = 4*.5 = 2 km
Thus total distance = 2+8 = 10.
This satisfy the last option - None of these.

Ravi since there are 4 options for the answer choice, is this a question from CAT material?
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Re: Interesting Time Distance Question  [#permalink]

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New post 30 Aug 2010, 08:12
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Since he reached 1 hr. ealier means that carriage met the guy 30 mins earlier than scheduled. That way the carriage saved a total of 1 hr (30 mins for each ways).

Since the man was supposed to meet the carriage after 1.5 hr, but he met the carriage 30 min earlier than scheduled , he walked for 60 mins before meeting the carriage. His speed was 4km/hr.
So, he walked for 4 km.

Distance from the house is 4(walked by the guy)+8(travelled using Carriage)=12km

Again, carriage had to travel 8km less (4km both way travelled by the man walking) which saved 1 hr, carriage's speed =8km/h

Hence, 1.

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Re: Interesting Time Distance Question  [#permalink]

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New post 30 Aug 2010, 11:15
Thanks Bunnel. Now I understand where I went wrong, but isn't it Very poorly written question.
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Re: A man arrives at a railway station 90mins before the time at  [#permalink]

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New post 03 Aug 2013, 14:12
A man arrives at a railway station 90mins before the time at which he had ordered his carriage to meet him. He walks at the rate of 4kmph and meets the carriage 8Km from his house and reaches 1 hour earlier than usual time. whats the speed of the carriage and distance from the station to his house.

The wording here is kind of ambiguous.

In one hour he travels 4km. In 1/2 the carriage travels the same distance. (it takes him 1 hour to travel 4km and the carriage travels at 8km/hr or 4km in one half hour) Therefore, the distance from the station to the house is 4+8=12 km. The speed of the carriage is 8km/hr.

ANSWER: A. 8kmph,12Km
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Re: A man arrives at a railway station 90mins before the time at  [#permalink]

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New post 12 Mar 2014, 02:39
and reaches 1 hour earlier than usual time.

What is "usual"?
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Re: A man arrives at a railway station 90mins before the time at  [#permalink]

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A man arrives at a railway station 90 minutes before the tim  [#permalink]

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New post 24 Sep 2014, 01:18
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Solving it mathematically,

Suppose the man travelled 'x' distance before meeting the carriage. Let the speed of carriage be 'S'.

Equation 1: Since the man arrived at the station 90 mins early therefore he should have reached his house 90 mins earlier if he would have traveled the whole distance by carriage, but he only reaches his house 60 mins earlier. This 30 mins delay is because of slower speed of walking which can be expressed as:
Time taken now: x/4+8/S
Time taken in normal case: (x+8)/S
therefore; (x/4+8/s) - (x+8)/S = 1/2 thus x(S-4)/4S = 1/2

Equation 2: Since carriage is running on time therefore it would have reached train station in time. The work done in 90 mins extra can be accounted by time spent for travelling 'x' distance by man + time that would have been spent had the carriage gone the whole distance to train station.
Therefore; x/4 +x/S = 3/2 thus x(S+4)/4S = 3/2

Dividing the two equation: (S-4)/(S+4) = 1/3 and thus S = 8km/hr
Substituting 'S' in one of the above equation, x = 4km and therefore total distance between train station and the house is 12km.

Therefore A option is correct.
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Re: A man arrives at a railway station 90 minutes before the tim  [#permalink]

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New post 15 Jun 2015, 03:29
RaviChandra wrote:
A man arrives at a railway station 90 minutes before the time at which he had ordered his carriage to meet him. He walks at the rate of 4 kmph and meets the carriage 8km from his house and reaches 1 hour earlier than usual time. What is the speed of the carriage and distance from the station to his house?

A. 8kmph,12Km
B. 12kmph,16km
C. 4kmph,12km
D. None of these

I used the substitution method to solve this problem.
answer is (D)
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A man arrives at a railway station 90 minutes before the tim  [#permalink]

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New post Updated on: 29 May 2017, 20:02
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A man arrives at a railway station 90 minutes before the time at which he had ordered his carriage to meet him. He walks at the rate of 4 kmph and meets the carriage 8km from his house and reaches 1 hour earlier than usual time. What is the speed of the carriage and distance from the station to his house?

A. 8kmph,12Km
B. 12kmph,16km
C. 4kmph,12km
D. None of these

plug-in approach:
s=speed of carriage
d=distance from station to house
d-8=distance man walks
2(d-8)/s=1 hour (saved)
plug options into 2d-s=16
only option A works

Originally posted by gracie on 05 Sep 2015, 16:15.
Last edited by gracie on 29 May 2017, 20:02, edited 1 time in total.
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A man arrives at a railway station 90 minutes before the tim  [#permalink]

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New post 29 May 2017, 10:35
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RaviChandra wrote:
A man arrives at a railway station 90 minutes before the time at which he had ordered his carriage to meet him. He walks at the rate of 4 kmph and meets the carriage 8km from his house and reaches 1 hour earlier than usual time. What is the speed of the carriage and distance from the station to his house?

A. 8kmph,12Km
B. 12kmph,16km
C. 4kmph,12km
D. None of these

The man started 90 minutes earlier than usual. Let us say he started at 12 instead of 1:30. The carriage should have met him before 1:30 because it usually reaches at that time but now since the man has already started walking, they should have met earlier. There is a saving of time for the carriage in the onward journey and in the return journey since the man walked some distance before the normal starting time. Since totally 60 minutes is saved which is the key, 30 minutes in the onward and 30 min in the return. The carriage should have met him 30 minutes before the normal meeting time of 1:30 and met him at 1. The man had walked from 12 to 1 i.e., 1 hr. His speed is 4km/hr . So the distance walked is 4 km. The same distance the carriage would have traveled in 30 minutes because we know it normally travels that distance also and it now saved 30 minutes. So the speed of the carriage is 4/0.5=8 km/hr. Total distance is 4+8=12km.
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A man arrives at a railway station 90 minutes before the tim   [#permalink] 29 May 2017, 10:35
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