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M11 #03

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M11 #03 [#permalink] New post 01 Oct 2008, 14:20
:shock: Would someone Kindly give me a step by step..GMAT for Dummies version...of this explanation? Thank you

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?

Choices: 50 kmh, 52 kmh, 56 kmh, 60 kmh, 70 kmh

Answer: Denote the speed of the oncoming train 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.

The correct answer is A.

[b]My Confusion is:

1) Again, please give me step by step on this.
2) I am under the impression that if you are measuring two objects moving in the same direction, you use the difference in their rates, not the total rate because the second train's tip will meet the first train's tip (meters apart) at it's own rate minus the first train's (also moving) rate.
3) Also, does the length of the passed train have any play?
4) The whole question and answer. Please help.
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Re: M 11 #3 [#permalink] New post 04 Oct 2008, 07:15
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Quote:
2) I am under the impression that if you are measuring two objects moving in the same direction, you use the difference in their rates, not the total rate because the second train's tip will meet the first train's tip (meters apart) at it's own rate minus the first train's (also moving) rate.


In the question, it says that the first train passes an oncoming train. So they're moving in opposite directions, not the same direction.

Quote:
3) Also, does the length of the passed train have any play?


Yes, otherwise you would have no way of calculating the speed.

Distance = Rate * Time, you need 2 of these to figure out the third. The problem gives you the time, 3 seconds, and asks for the rate/speed whatever you want to call it. If they didn't give you the distance, you wouldn't be able to calculate the answer.

Maybe it'll be easier to eliminate the speed of the first train. Someone is sitting still, and a 75 meter train passes them. Three seconds pass between the time the front of the train passes the observation point and the time the end of the train passes the observation point. When the end of the train is at the observation point, the front of the train is 75 meters down the track. So the front of the train traveled 75 meters 3 seconds. 75/3 = 25 meters per second which, as it says in the answer is equivalent to 90 kmh. Just memorize this and other common conversions so you don't have to actually calculate them at test time.

Now let's go back to the original problem where we're on a moving train. As I cleared up above, the trains are moving towards each other so if the speed of train 1 is v1 and the speed of train 2 is v2, they pass each other at a combined speed of v1 + v2. We've already calculated this combined speed, which is the speed that will be apparent to a passenger on either train, in the previous paragraph, 90 kmh. The speed of train 1 is given in the problem as 40 kmh.

So solve 40kmh + v2 = 90km for v2
v2 = 50kmh which is the answer to the problem.
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Re: M 11 #3 [#permalink] New post 04 Oct 2008, 15:25
Thank you csvobo. I'm not sure Why I assumed "oncoming" meant a train that was "upcoming" on the other train and that the passenger turned right to look towards the back of the train. Wow, I need to pay more attention to these words. Thank you!
Re: M 11 #3   [#permalink] 04 Oct 2008, 15:25
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M11 #03

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