A train crosses a bridge of length 500 m in 40 seconds and a lamp post

Length of train = L
Speed of train = S

(L+500) = S*40
L = S*15 => S = (L / 15)

=> L+500 = (L / 15) * 40
=> L = 300m

Option E

let the length of the train is L
L/15 = (L+500)/40
solving, L=300

IMO: E

Speed = \(\frac{Distance}{time}\)

Let Length of train be \(L_T\)
Let Length of Bridge be \(L_B\)
Speed of train be S

While crossing Lamp post
S = \(\frac{L_T}{15}\)
\(L_T\) = 15S --(i)

While crossing Bridge
s = \(\frac{L_T + L_B}{40}\)
40s = \(L_T\)+\(L_B\)
40s = 15s+500 (from eq--(i) & given length of bridge = 500)
s= 20

Sub in eq-(i) we get \(L_T\) = 300

You have two concepts covered in this topic: a train which crosses a stationary object with (1) and without (2) length:

Crossing a stationary object with length (bridge = 500m):

r * t = Length of Train + Length of object ->> Where r = spdeed and t = time

In this question we get: 40r = 500 + Length of Bridge

then crossing a stationary object without length/with negligible length:

r*t = Length of train

So we have 15r = Length of Train ->> plug in above to get:

40r = 500 + 15r
25r = 500
r = 20

Plug in r = 20 in 15r = Length of train to get Length = 300.

We can let x = the length of the train in metres. To cross the 500-m bridge in 40 seconds, the head of the train must enter at one end of the bridge and the tail of the train must exit from the other end of the bridge. Therefore, in 40 seconds, the train actually travels the length of the bridge plus the train’s body length. In other words, the train travels (500 + x) metres in 40 seconds. If we let r be the speed of the train, we have:

500 + x = 40r

Since there are two unknowns, we need another equation in order to solve for x. We will use the fact that the train crosses a lamp post on the bridge in 15 seconds. This means that after the head of the train first crosses the lamp post, it takes 15 seconds for the train’s tail to cross the lamp post too. In other words, the train travels a distance equal to its body length in 15 seconds. So, we have:

x = 15r

Substitute x = 15r in the first equation and we have:

500 + 15r = 40r

500 = 25r

20 = r

So, we know the speed of the train is 20 m/s and its body length is x = 15r = 15(20) = 300 metres.

Answer: E

hey GMATinsight
not familiar with such type of questions. are there specific formulas for such concepts ?
thanks!

hey , this should help with train related concepts https://www.hitbullseye.com/Quant/Probl ... Trains.php

Solution

Given:

• A train crosses a bridge of length 500 m in 40 seconds.
• A train crosses a lamp post on the bridge in 15 seconds.

To find:

• Length of the train in metres.

Approach and Working

Let the length of train is ‘t’ metres and speed of train is ‘s’ m/s.

• Therefore, we can frame two equations.

A train crosses a bridge of length 500 m in 40 seconds.

• In this case, train covers its length as well as the distance of bridge.

o Hence, t+500= 40*s------(1)

A train crosses a lamp post on the bridge in 15 seconds.

• In this case, train covers only its length.

o Hence, t= 15*s

• Substituting this value of t in equation 1, we get:

o 15s +500 = 40s
o 25s=500
o s= 20 m/s

 Hence, t= 15*20= 300 m

Correct answer: Option E

dave13

Search for Time speed and distance questions and you would find plenty of such questions

There are not many formulas only a few concepts like Relative speed, River and boat problems etc.

I'm not able to understand the basic thing. Can you please explain why we add length of train and bridge length but in case of lamp post, why do we take train length? Shouldn't it be lamp post's length?

\(\frac{500+x}{40}\) = \(\frac{x}{15}\)

\(\frac{t + 500}{40} = \frac{t}{15} \) {Speed of train is constant}

So, \(15t + 7500 = 40t\)

Or, \(25t = 7500\)

Or, \(t = 300\), Answer will be (E)

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