Formula for finding out problems on train

1. km/hr to m/s conversion: a km/hr = (a*(5/18))m/s.

2. m/s to km/hr conversion: a m/s = (a*(18/5))km/hr.

3. Time taken by a train of length l metres to pass a pole or standing man or

a signal post is equal to the time taken by the train to cover l metres.

4. Time taken by a train of length l metres to pass a stationery object of

length bmetres is the time taken by the train to cover (l + b) metres.

5. Suppose two trains or two objects bodies are moving in the same direction

at u m/s and v m/s, where u > v, then their relative speed is = (u - v) m/s.

6. Suppose two trains or two objects bodies are moving in opposite directions

at u m/s and v m/s, then their relative speed is = (u + v) m/s.

7. If two trains of length a metres and b metres are moving in opposite

directions at um/s and v m/s, then:

The time taken by the trains to cross each other = (a + b) sec.

(u + v)

8. If two trains of length a metres and b metres are moving in the same

direction at um/s and v m/s, then:

The time taken by the faster train to cross the slower train = (a + b) sec.

(u - v)

9. If two trains (or bodies) start at the same time from points A and B

towards each other and after crossing they take a and b sec in reaching B and

A respectively, then:

(A's speed) : (B's speed) = (b : a)

10. Speed = distance/time s = d/t

11. velocity = displacement/time v = d/t.

12. Time taken by a train x meters long to pass a pole or standing man or a

post

= Time taken by the train to travel x meters.

13. Suppose two trains or two objects are moving in the same direction at v1

m/s and v2 m/s where v1 > v2,

then their relative speed = (v1 v2) m/s

14. Suppose two trains or two objects are moving in opposite directions at v1

m/s and v2 m/s ,

then their relative speed = (v1+ v2) m/s

15. Assume two trains of length x metres and y metres are moving in opposite

directions at v1 m/s and v2 m/s, Then

The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds

16. 10. Assume two trains of length x metres and y metres are moving in the

same direction at at v1 m/s and v2 m/s where v1 > v2, Then

The time taken by the faster train to cross the slower train = (x+y) / (v1-

v2) seconds

17. Suppose 2 Trains begins at the same time from X and Y towards each other

and after crossing they take x & y secs in reaching Y and X correspondingly,

then:

(Xs speed) : (Ys speed) = SQRT(y) : SQRT(x)

18. Rule 1: When two trains are moving in opposite diections, then relatve

speed will be the addition of their individiual speeds.

19. Rule 2: When two trains are moving in same diection, then relatve speed

will be the subtrction of their individiual speeds.

20. Rule 3: On passing a platform by a certain train the net distance

travelled is the sum of length of train and the length of platform both.

21. Rule 4: When a train passes through a pole or person standing, net

distance travelled to pass is the length of the train

Sample Example

Ex

A train 120 m long is running at the speed of 54 km /hr. Find the time taken it to pass a man standing near the railway track ?

A

speed of train = [54 * ( 5 / 18 ) ] = 15 m / sec

length of train = 120 m , So required time :

Time taken = (120/15) = 8 sec.

Ex

train is moving at a speed of 54 km / hr. If the length of the train is 100 meters, how long will it take to cross a railway platform 110 meters long ?

A

speed of train = [54 * ( 5 / 18 ) ] = 15 m / sec

Distance covered in passing the platform = 100 + 110 = 210 m

therefore Time taken = (210/15) = 14 sec.

Ex

Two trains 125 m and 100 m in length respectively are running in opposite directions, one at the rate of 50 km / hr and the other at the rate of 40 km /hr. At what time they will clear each other from the moment they meet ?

A

Relative speed of trains = (50 + 40) km / hr = [90 * ( 5 / 18 ) ] = 25 m /

sec

Total length to be travelled = 125 + 100 = 225 m

therefore Time taken = (225/25) = 9 sec.

Ex

Two trains 110 m and 100 m in length respectively are running in same directions, one at the rate of 100 km / hr and the other at the rate of 64 km / hr. At what time faster train will clear other train from the moment they meet ?

A

Since trains are running in same direction, so relative speed = 100-64 = 36

km / hr = [ 36 * ( 5 / 18 )] = 10 m / sec

Total length to be travelled = 110 + 100 = 210 m

therefore Time taken = (210/10) = 21 sec.

Ex

A train passes a standing pole on the platform in 5 seconds and passes the platform completely in 20 seconds. If the length of the platform is 225 meters. Then find the length of the train ?

A

Let the length of the train is x meter

So speed of train =( x / 5 ) m / sec

Also speed of train = ( 225 + x ) / 20 m/sec

x/5 = (225+x)/20, therfore x = 75 m.

Ex

Two trains of length 115 m and 110 m respectively run on parallel rails. When running in the same direction the faster train passes the slower one in 25 seconds, but when they are running in opposite directions with the same speeds as earlier, they pass each other in 5 seconds. Find the speed of each train ?

A

Let the speed of trains be x m/sec and y m/sec espectively.

When they move in same direction their relative speed is : x - y

When they moves in opposite direction their relative speed is : x + y

x-y = (115+110)/25, x+y = (115+110)/5

On solving two equations x=27 m/s and y=18 m/sec

Ex

The product of two numbers is 60480 and their HCF is 12 . Find the numbers ?

A

Since HCF s 12, so the two numbers will be multiple of their HCF

let the first number is 12P and the second number be 12Q

12P 12Q = 60480

P Q = 420

Now pair of numbers whose product is 420 is

( 420, 1 ) ( 210, 2 ) ( 140, 3 ) ( 105, 4 ) ( 60, 7 )

( 20, 21 )

Out of these ( 210, 2 ) is not prime so neglected

Now the required numbers will be ( 42012, 112 ) ( 14012, 312 ) (

10512, 412) ( 60 12 , 712 ) ( 2012, 2112 )

( 5040, 12 ) (1680, 36) ( 1260, 48) ( 720, 84 ) ( 240, 252 ) be the

required numbers

Ex

Find the greatest number that will divide 37, 109 and 157 so as to leave the same remainder in each case ?

A

Let the remainder be x, then the numbers :

( 37 - x ) ( 109 - x ) ( 157 - x ) must be divisible by the

required number.

Also if two numbers are divisible by the certain number then their

difference is also divisible by that number

( 109 - x ) - ( 37 - x ) = 72

( 157 - x ) - ( 109 - x ) = 48

( 157 - x ) - ( 37 - x ) = 120

So, the numbers 72, 48, 120 will also be divisible by that number, So HCF

of 72, 48, 120 is 24, therefore required number will be 24

Formula for finding out problems on trainSample ExampleA train 120 m long is running at the speed of 54 km /hr. Find the time taken it to pass a man standing near the railway track ?train is moving at a speed of 54 km / hr. If the length of the train is 100 meters, how long will it take to cross a railway platform 110 meters long ?Two trains 125 m and 100 m in length respectively are running in opposite directions, one at the rate of 50 km / hr and the other at the rate of 40 km /hr. At what time they will clear each other from the moment they meet ?Two trains 110 m and 100 m in length respectively are running in same directions, one at the rate of 100 km / hr and the other at the rate of 64 km / hr. At what time faster train will clear other train from the moment they meet ?A train passes a standing pole on the platform in 5 seconds and passes the platform completely in 20 seconds. If the length of the platform is 225 meters. Then find the length of the train ?Two trains of length 115 m and 110 m respectively run on parallel rails. When running in the same direction the faster train passes the slower one in 25 seconds, but when they are running in opposite directions with the same speeds as earlier, they pa...The product of two numbers is 60480 and their HCF is 12 . Find the numbers ?Find the greatest number that will divide 37, 109 and 157 so as to leave the same remainder in each case ?