A train of length L is traveling at a constant velocity and passes a

Speed = L/t = (L+P)/3t. So P = 2L

The distance the train is traveling isn't just L, it's L plus the length of the pole. This question assumes that the length of the pole zero.

Length of pole is assumed 0. From case 1 we get the speed of the train =L/t
In case 2 we need distance =speed x time =L/t x 3t=3L, but this 3L is the total distance inclusive of the length of the train which is L. so length of the platform =3L-L=2L.
Hope this explanation helps.

Hi All,

This question can be solved by TESTing VALUES.

We're told that a train has a length of L and passes a pole in T seconds.

Let's TEST:
L = 3 (feet)
T = 2 (seconds)

Since the pole is NOT given a length, we have to assume that that value is negligible (and does NOT factor into the calculations).

So, the entire length of the train passes the pole in 2 seconds. Thus, it's traveling 3 feet every 2 seconds.

Next, we're told that this train passes a platform in 3T seconds. This means that the train travels 3(2) = 6 seconds and travels 3(3) = 9 feet.

Those 9 feet are made up of the train (3 feet) and the platform (9 - 3 = 6 feet).

The question asks us for the length of the PLATFORM. Using these VALUES, we know that the PLATFORM = 6 when L = 3.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

let r=rate
let p=length of platform
L+p=(3t)(r)=3rt
(L+p)/3=rt
L=rt
(L+p)/3=L
L+p=3L
p=2L

Look at the question logically:

"A train of length L is traveling at a constant velocity and passes a pole in t seconds"
To cross a pole (which can be considered a point), why does the train take t secs? Because the whole length of the train has to cross it.

" If the same train travelling at the same velocity passes a platform in 3t seconds"
Why does the train take extra 2t secs now? Because of the length of the platform. So the length of the platform must be 2L.

Answer (D)

It's a very simple question..
So, Train of length L passes a pole in t seconds. Velocity of pole v= L/t

Now this train of length L passes platform of say length p passes in 3t seconds .
So, (L+p)= v (3t) = L/t(3t) 3L
-> p = 2L

Answer D

We can also analyse it and solve it in a very short span of time as VeritasPrepKarishma suggested...

Train of length L passes a pole in t seconds i.e. it covers L in t seconds.
So, in 3t seconds it covers 3L distance, out of which L is length of train and 2L is length of platform... very simple, isn't it? :)

We can let p = the length of the platform in meters and r = the rate the train is traveling.

When we say the train crosses a platform in 3t seconds, it really means it takes 3t seconds for the nose of the train to enter one end of the platform and the tail of the train to exit the other end of the platform. Thus, in 3t seconds, not only does the train travel the entire length of the platform but also it travels its body length L. Thus, we have (using time x rate = distance formula):

3t * r = p + L

We are also given that the train crosses a pole (notice that the pole has a negligible width) in t seconds. So when the train crosses the pole, it only travels its body length in t seconds. Thus we have:

t * r = L

Subtracting these two equations, we have:

2tr = p

Since L = tr, and p = 2tr, then p = 2L.

Answer: D

Interesting question. One needs to pay attention to details. First, the train crossed a "pole" and the second time the train crossed the "platform". In the former case, assume the length of the pole is a limit tending to zero vis-a-vis the length of the train. Once you realize this you will not fall for the trap answer and realize the train is traveling a 2L distance which is the length of the platform.

Train passes a pole in T seconds.

In order to pass the pole, the train needs to put its entire length of L from the head of the train to the back of the train in front of the pole. This distance covered will = length of train = L

Speed of Train = (L meters) / (T seconds)


In order to pass the platform, once the head of train reaches the beginning of the platform, the train will need to cover:

distance of the platform (let it be X)

+

put its entire distance in front of the platform (another distance = Length of Train = L)

To pass the platform, the train will therefore need to cover a distance = L + X

where X = length of platform.

The time taken to do this is = 3T seconds


Speed * Time = Distance of Train and Platform

(L/T) * (3T) = L + X

3L = L + X

X = 2L = length of platform

D

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Given: A train of length L is traveling at a constant velocity and passes a pole in t seconds.
Asked: If the same train travelling at the same velocity passes a platform in 3t seconds then what is the length of the platform:

velocity of the train = L/t m/s
Let the length of the platform be x m.
Distance travelled = L/t * 3t = 3L = L + x
x = 2L

IMO D