**Main-Topic:** Time, Speed Distance (TSD)

**Subtopic:** Trains and Platforms.

you’ll be given some data and asked to find the length of railway platform or speed of train etc.

- Visualizing the Concept: Train and Pen
- Case 1: One Train One Man (or Tree, Pole)
- Case 2: One Train One Platform (or bridge, tunnel)
- Case 3: Two Trains
- Unit Conversions
- Previous Articles on Aptitude

# Visulize: Pen and Train

Before going into Trains concept, try to visulize following. (ofcourse technically incorrect but for concept clarity)

You must have seen the advertizement of a ballpen “Flair Writometer”. They claim it can write for 2,500 metres and even provide a marking scale on the ballpen refill to measure the ‘distance’! See the photo

So You’re standing on a Railway platform, holding that writometer pen in your hand, pointing it straight towards the upcoming train’s side, like this photo

The train passes, and your pen makes a straight line on the entire body of this train.

Now you open the pen’s refill and check the markings, obviously few mili-litres of ink is used and marking would show that you’ve drawn a line of 200 meters.

Meaning length of this train is 200 m.

At the same time, your friend was sitting in the window seat of the same railway and he too held a writometer pen in his hand pointing it out of the window and towards the platform.So as the train passed, his pen also drew a straight line on the entire platform. Later he checks the refill of his pen: it drew a line of 1000 meters.

Meaning length of this railway platform is 1000 m.

so what is the total distance or length of lines drawn by the two pens?

200m of train + 1000m of platform = 1200m.

Now questions

# Case1: one men, one train.

A railway train running at the speed of 20 meter per second, crosses a man in 10 seconds. What is the length of the train?

For all the problems about train, boats, cars, scooters, men required to complete a job, tata nano and kingfisher airplane, in short TSDW (time,speed, distance, work) you need to remember only one “STD” formula

ST=D

speed (s) x time (t) = distance (d)

Back to the question, what is given?

Visualise the situation: you are the man, holding the writometer pen and the train crossed you in 10 seconds. What is the distance of the line drawn by your pen on the body of the train?

Speed (s)= 20 m/s (given)

Time (t)= 10 seconds (given)

so

s x t = d

20 x 10=200 m.

final answer: length of the train is 200 m.

## Alternatives:

Instead of man, they can rephrase and ask: Train crossed a “Tree, Telephone-pole, Tiger, Kalmadi, Raja..” The concept remains the same. Everybody is holding a pen.

# Case2: one platform, one train.

Question: a train running at the speed of 20 m/s, crosses a railway platform in 20 seconds. If the length of the train is 100 m, what is the length of this railway platform?

So in this case, you were standing on a platform holding a **blue** pen and I am sitting in the railway also holding a **purple** pen.

What is the total length of lines drawn by both of us?

Total Length (or distance)= Your **blue** pen drew 100 m line + My **purple** pen drew (x) m line.

: The total of **blue** and purple color line in above photo

So total distance covered by the train is (100+x)

Now ,

speed is given: 20m/s

time is also given: 20 seconds

So apply the STD formula

s x t = D

20ms x 20 seconds = (100+x)

400m = 100+x

x=400-100

x=300m

Final answer: length of this platform is 300 metres.

Alternative: instead of railway platform, they can use the word “Bridge, Tunnel” but the principle remains the same.

# Case3: two trains crossing each other

**Two trains are coming from opposite directions. Both of them are running at the same speed 20 meter per second.**

It takes 20 seconds for them to completely cross each other. If the length of first train is 90 m, what is the length of the second train?

Visualise: you are holding a pen sitting in first train, I am holding a pen sitting in the second.

The total length of line drawn is 90+x m

Since both of the trains are moving, and moving in opposite directions. Meaning it will take less time to draw these lines.

Hence

Final speed (s)

= speed of first train+ speed of second train.

= 20+20

= 40 m/s

Now it can be solved just like the case #2.

In the case #1 and #2, Man and Platform were stationary (unmoving) objects so their speed was zero, hence we took train’s speed as our final speed. But here second train is also moving so we’ve to consider it.

## Alternative:

same question, but instead of two trains , it is one man riding on a bike and another train coming from the opposite direction.

Here’s the length of length of line

=Line drawn on Train + Line drawn on Bike

= 90m +zero meter. (because length of bike is negligible, we take it zero)

So Distance (d)= 90.

but for the speed we have to add

=Train + Bike

=20+20

S=40 m/s

After that, principle remains the same as Case #3

# Unit conversions

It is essential that before applying the STD formula, all the unit is must be in the same system, else you will get incorrect answer.

and they will give that incorrect answer in one of the four options [a/b/c/d] so you will confidently tick it but end up loosing in negative marking.

So in the STD formula. Everything must be

either in meter,second,

or in kilometer, hour.

Remember

# 1 km per hour = 5/18 meter per second.

## How did we derive this?

1 kilometer per hour

=1000 meters per 60 minutes

=1000 meters per (60×60) seconds

=1000 meters per 3600 seconds

=10 meters per 36 seconds (cut the zeros)

=5 meters per 18 seconds. (half of 10 and 36)

Meaning

if a train is running a 36 kms per hour then, its speed in meter per second is:

1 kmph = 5/18 ms

36 kmph= 36x (5/18) = 10 m/s.

If the reverse is asked, (10 m/s then how much in kmph?),

simply multiply with reverse 18/5, because

5/18 ms = 1 kmph

1 ms= 18/5 kmph

10 ms= 10 x (18/5) kmph = 36 kmph.

## 72 Comments on “[Aptitude] Trains, Platforms TSD (Time,Speed,Distance) made easy”

2sin(πx/2)=x2 +1/x2

Then what is value of the (x – 1/x)

-2sec^2(πx/2)

if two trains are crossing each other and plateform length is given like 1 km and train’ s length is not mentioned and speed for both trains are as 40 km/ hr and 50 km/ hr respectively..could anyone please tell me it’ s valid answer which can satisfy all the conditions

Local trains leave from a station at an interval of 15minutes at a speed of 16km/hr.A man moving from opposite side meets the trains at an interval of 12 minutes.The speed of the man is :

Plz anyone tell me the solution

could you plz send me the any imp topic material sruthi..

A goods trains of length 600 meters traveling at a uniform speed completely passes the platform of a station 45 seconds. If the

length of platforms is 300 meters. Find the speed of the train in kilometers per hour?

given that length of both the train and platform are 600 and 300m.

so the distance is 600+300 =900m

train passes platform 45sec. so time also given, we have to find speed of train in km/hr?

speed =distance/time

s =900m/45sec = 20m/sec

convert into km/hr so multiply by 18/5

20*18/5 =72km/hr is the answer……

Sir, How to solve these using STD table?

A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?

A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is?

a train is running with d speed of 93km,another train with speed of 51 km coimng from opposite side and crosses each other in 24 seconds. the train crosses a bridge in 66 seconds then length of bridge in meters is?

plz solve dis…

My first visit to mrunal.org and I found it so damn impressive.

a person going in a cae at 30kmph reches t1minutes late, if he goes at 40kmph he reaches the office t2 minutes early. find the distance between the office and his house

hi sir or medam

any body can you plz send me all type of rrb material or railway jobs matreial on most important topics..

on given my below id.

at least few material also send me sir..or medam

एक ट्रेन x से शहर y की तरफ़ जाते हुए 1 घंटे बाद दुर्घटना ग्रस्त हो जाने पर इसे आधा घंटा रोक लिया जाता है. इसके बाद यह अपनी सामान्य चाल की 4/5 चाल से चलकर y पर 2 घंटे देरी सि पहुँचती है. यदि दुर्घटना से पहले 80 km चली होती तो केवल 1 घंटे. देरी से पहुँचती.दुर्घटना सि पहले ट्रेन की चाल बताओ.तथा उंकि बीच की दूरी क्या है

Yes I too got 710.

A train, 300m long, passed a man, walking along the line in the same direction at the rate of 3 kmph in 33 seconds. The speed of the train is:

please solve it

Plz provide pictures they are not working

A train after travelling 50 km from A meets with an accident and proceeds at 4/5th of former speed and reach B 45 mins late.Had accident happened 20 km further on,it would arrive 12 min sooner.Find original speed and distance?

Can u please suggest a good book for quant

The driver of a car driving @376km/hr locates a bus 40meters ahead of him .after 20secs the bus is 60meters behind .the speed of the bus is.

The two train are travel in the same direction at 56K.M and 29K.M an hour and the faster train passes a man in the slower train in 16 seconds. Find the lenth of faster train.

A. 120M

B. 150M

C. 140M

D. 100M.

Please tell me how did we solve it

relative speed 56-29 = 27kmph = 7.5mps( so man is static now= take him as a point)

train crosses that point in 16 sec with speed 7.5 mps , so distance traveled by train is it’s length

d=sxt = 7.5*16=120meter