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D01-25

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D01-25 [#permalink]

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A train is traveling at a constant speed from city A to city B. Along this trip the train makes three one-hour stops and reaches city B. At city B the train is stopped again for 1 hour. After that the train makes the return trip from city B to city A at a constant speed which is twice the speed of the first trip. Along this return trip the train makes ten thirty minutes stops and reaches city A. If both trips took the same amount of time, how many hours was the roundtrip?

A. 14
B. 15
C. 16
D. 17
E. 18
[Reveal] Spoiler: OA

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Official Solution:

A train is traveling at a constant speed from city A to city B. Along this trip the train makes three one-hour stops and reaches city B. At city B the train is stopped again for 1 hour. After that the train makes the return trip from city B to city A at a constant speed which is twice the speed of the first trip. Along this return trip the train makes ten thirty minutes stops and reaches city A. If both trips took the same amount of time, how many hours was the roundtrip?

A. 14
B. 15
C. 16
D. 17
E. 18


Let the time for each trip be \(t\) hours, then the roundtrip took \(t+t+1=2t+1\) hours (each trip took the same amount of time plus 1 hour of waiting after the first half of the trip at city B).

Next, during the first trip the train was stopped for \(3*1=3\) hours and for the second trip it was stopped for \(10*0.5=5\) hours, so for the first trip it was moving at some constant rate, say \(r\), for \(t-3\) hours and for the second trip it was moving for the constant rate of \(2r\), for \(t-5\) hours.

Now, since the distance for the first and the second trip was the same then \(r*(t-3)=2r*(t-5)\), which gives \(t=7\).

So, the roundtrip took \(2t+1=15\) hours.


Answer: B
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D01-25 [#permalink]

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New post 31 Dec 2014, 01:19
No hours stopped from A to B = 3 hrs
No hours stopped from B to A = 5 hrs (10*0.5)
No hours stopped at B = 1 hour

Let Distance be D
Speed traveled by train from A to B = x
Speed traveled by train from B to A = 2x
3+(D/X)=5+(D/2x) (given)
D = 4x ------------------(1)

Total time = Time taken from AtoB + Time taken from BtoA + hours stopped in B
Total time = 3+(D/x) + 5+(D/2x) + 1
Sub (1) in above

total time = 15hrs Option (B)

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Re: D01-25 [#permalink]

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New post 15 Sep 2015, 01:31
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My solution:
Let D be the distance and r be the rate:
As given: D/r + 3 (three one-hour stops) = D/2r + 5 ( ten thirty minutes stops = 5 hour)
=> D/r - D/2r = 2
=> D=4r => 4r/r + 3 = 7
=> Total time: 7+7+1=15
Is it right!

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Re: D01-25 [#permalink]

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New post 29 Dec 2015, 03:03
Hello Bunuel,

Could you please explain why we are not considering the total waiting time (during the trip) in total trip time?

Thanks

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Re: D01-25 [#permalink]

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New post 03 Oct 2016, 14:13
hi guys i have a doubt,

question says
A train is traveling at a constant speed from city A to city B. Along this trip the train makes three one-hour stops and reaches city B. At city B the train is stopped again for 1 hour.

now it means 3 hours+1 hours at station B?

now i was able to find running time as 7 hours

but than 7+5(stoppage time for B---------->A)+3(stoppage time for A-------->B)+1 hour for station B
that gives us 16 hours

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Re: D01-25 [#permalink]

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New post 25 Dec 2016, 12:22
shubham2312 wrote:
hi guys i have a doubt,

question says
A train is traveling at a constant speed from city A to city B. Along this trip the train makes three one-hour stops and reaches city B. At city B the train is stopped again for 1 hour.

now it means 3 hours+1 hours at station B?

now i was able to find running time as 7 hours

but than 7+5(stoppage time for B---------->A)+3(stoppage time for A-------->B)+1 hour for station B
that gives us 16 hours


Hi,

How did you find the running time as 7 hours ? What did you assume ?
I guess based on how you got the running time as 7 hours, all can look out for the mistake that you are making.
In the solution posted by Bunuel, it is very evident that 7 hours is the total time for trip between A->B (7 hours include the stops that the train made)
Thanks !

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Re: D01-25 [#permalink]

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New post 25 Oct 2017, 12:50
Let's say that the trip on the way there took time T. Since distance is constant and we increase the speed two times, then we know that the time is cut in half. so T is the time to get there and .5T is the time to get back without considering any of the stops. Since we know that roundtrip is the same, then T + 3 = .5T + 5.

Solve for T = 4. so 4+3+ .5(5) + 5 + 1 (the extra hour one stayed).

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Re: D01-25   [#permalink] 25 Oct 2017, 12:50
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