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Re: Interesting Time Distance Question [#permalink]
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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.

Thanks
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Re: Interesting Time Distance Question [#permalink]
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]
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]
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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RG800 wrote:
and reaches 1 hour earlier than usual time.

What is "usual"?


Usual time = by schedule, so the time the man reaches home when the train arrives by schedule and not 90 minutes earlier.

Hope it's clear.
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A man arrives at a railway station 90 minutes before the tim [#permalink]
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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]
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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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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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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