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M11-03

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M11-03  [#permalink]

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New post 16 Sep 2014, 00:43
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  35% (medium)

Question Stats:

81% (01:33) correct 19% (02:42) wrong based on 99 sessions

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New post 16 Sep 2014, 00:43
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Official Solution:

If a passenger sitting near the window in a train moving at 40 kmh noticed that it took 3 seconds for the oncoming train to pass by, what was the speed of the oncoming train if the length of the oncoming train was 75 meters?

A. 50 kmh
B. 52 kmh
C. 56 kmh
D. 60 kmh
E. 70 kmh

Denote the speed of the oncoming train as \(V\). Then its speed relative to the passenger is \(V + 40\). 75 meters in 3 seconds is the same as 25 meters in 1 second or 90 kmh. Thus, \(V + 40 = 90\) and \(V = 50\) kmh.

Answer: A
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New post 25 Jan 2015, 12:18
Bunuel wrote:
Official Solution:

If a passenger sitting near the window in a train moving at 40 kmh noticed that it took 3 seconds for the oncoming train to pass by, what was the speed of the oncoming train if the length of the oncoming train was 75 meters?

A. 50 kmh
B. 52 kmh
C. 56 kmh
D. 60 kmh
E. 70 kmh

Denote the speed of the oncoming train as \(V\). Then its speed relative to the passenger is \(V + 40\). 75 meters in 3 seconds is the same as 25 meters in 1 second or 90 kmh. Thus, \(V + 40 = 90\) and \(V = 50\) kmh.

Answer: A


"Then its speed relative to the passenger is \(V + 40\)." ----> Could not understand it. Would you please clarify this a little bit.

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New post 26 Jan 2015, 03:29
Mahmud6 wrote:
Bunuel wrote:
Official Solution:

If a passenger sitting near the window in a train moving at 40 kmh noticed that it took 3 seconds for the oncoming train to pass by, what was the speed of the oncoming train if the length of the oncoming train was 75 meters?

A. 50 kmh
B. 52 kmh
C. 56 kmh
D. 60 kmh
E. 70 kmh

Denote the speed of the oncoming train as \(V\). Then its speed relative to the passenger is \(V + 40\). 75 meters in 3 seconds is the same as 25 meters in 1 second or 90 kmh. Thus, \(V + 40 = 90\) and \(V = 50\) kmh.

Answer: A


"Then its speed relative to the passenger is \(V + 40\)." ----> Could not understand it. Would you please clarify this a little bit.


This is the concept of Relative Speed.

When two objects (speeds V1 and V2) move in opposite directions (towards each other or away from each other), they cover the distance between them (or create distance between them) at the rate of (V1 + V2).

When two objects move in same direction, their speeds get subtracted.

Theory on Distance/Rate Problems: distance-speed-time-word-problems-made-easy-87481.html

All DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
All PS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=64

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Re: M11-03  [#permalink]

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New post 27 Jan 2015, 09:36
Bunuel wrote:
Mahmud6 wrote:
Bunuel wrote:
Official Solution:

If a passenger sitting near the window in a train moving at 40 kmh noticed that it took 3 seconds for the oncoming train to pass by, what was the speed of the oncoming train if the length of the oncoming train was 75 meters?

A. 50 kmh
B. 52 kmh
C. 56 kmh
D. 60 kmh
E. 70 kmh

Denote the speed of the oncoming train as \(V\). Then its speed relative to the passenger is \(V + 40\). 75 meters in 3 seconds is the same as 25 meters in 1 second or 90 kmh. Thus, \(V + 40 = 90\) and \(V = 50\) kmh.

Answer: A


"Then its speed relative to the passenger is \(V + 40\)." ----> Could not understand it. Would you please clarify this a little bit.


This is the concept of Relative Speed.

When two objects (speeds V1 and V2) move in opposite directions (towards each other or away from each other), they cover the distance between them (or create distance between them) at the rate of (V1 + V2).

When two objects move in same direction, their speeds get subtracted.

Theory on Distance/Rate Problems: distance-speed-time-word-problems-made-easy-87481.html

All DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
All PS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=64


Hi Bunuel,

Thanks a million to reply. I understood the concept of relative speed. However, I have a query.
Why '75 meters in 3 seconds'? Why not '75 meters plus passenger's train length in 3 seconds'?

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New post 27 Jan 2015, 09:40
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Re: M11-03  [#permalink]

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New post 13 Dec 2015, 15:49
"75 meters in 3 seconds is the same as 25 meters in 1 second or 90 kmh" How is the 90 kmh derived from this?
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Re: M11-03  [#permalink]

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New post 13 Dec 2015, 16:02
LostinNY wrote:
"75 meters in 3 seconds is the same as 25 meters in 1 second or 90 kmh" How is the 90 kmh derived from this?


I would guess 3600*25 (3600 seconds in one hour). Alternatively if you know the shortcut (which you definitely do not need to know for the GMAT), 10 m/s equals to 36km/h, so 25m/s = 36*2.5=90
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New post 22 Aug 2016, 06:06
I think this is a high-quality question and I agree with explanation.
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Re M11-03  [#permalink]

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New post 22 Aug 2016, 06:12
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Your explanation is short and superb. However, i equated the distance covered by two trains during the 3 second like:

40 * (3/3600) = x + 75/1000

where 40 is the speed of the passenger's train, 3/3600 is time 3 seconds in hours , x is the distance travelled in these 3 seconds and 75 meters expressed in KM. I got x = 1/24 KM. Speed of oncoming train = Distance / Time = (1/24) / (3/3600) which gave me 50KM/Hr. Is this approach correct?
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Re: M11-03  [#permalink]

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New post 30 Dec 2018, 11:03
I still did not get this part:

V+40=90V

WHY IS IT SO?
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Re: M11-03   [#permalink] 30 Dec 2018, 11:03
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