Aptitude - Problems on Trains - Discussion

@Priya,

Speed=Distance/Time.
here Distance=length of train + length of bridge.

in this question we have no idea about the length of train and speed of the train.
so let the length of the train be x metres and speed be y m/sec.
then speed of train y = (x+264)/20.

we can't solve because vale of x and y is unknown.
now speed of the train relative to telegraph post
i.e. y = x/8.
comparing the two above equation
x/8 = (x+264)/20
3x = 528
x = 176 which is length of train.
now put the value of x in any of the above two equation.
therefor speed y = 176/8
y = 176/8 *18/5 (because to convert in kmph)
y = 79.2 km/hr

Simple answer!

A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?

Train moves past a Tpost in 8 sec, 264 m long bridge in 20 sec (difference is 12 sec) now calculate the length of the train = 8 x 264 / 12 = 178. mtrs.

Formula -> Time= distance/speed
=> 8 = 178/speed (178/8 = 22 meters/seconds).
Now convert 22 meter/sec to Km/hr by multiplying 22 x 18/5 or (3600 sec/1000 meters) = 79.2.

Here we don't know the speed of train, So let x be the speed of train , According to the formula t = a+b/u+v.

1st case where train crosses the platform,

let b be the length of platform.
t=300+b/x+0 => 39=300+b/x => 39x=300+b ---- eqn 1.

2nd case where train passes the pole,
t=300+0/x+0 =>18=300/x =>x=300/18 ----eqn 2.

We go value for x. So substitute in equation 1.

39*300/18=300+b.

650-300=b.

b=350----->length of platform.

Hi, everyone.

If a train pass any object whose length is negligible (ex:- man, post) we don't need to add d length of d train and d object, so we use speed= (distance/time) formula. Time is specified so (x/y) =8 ->1. If a train passes any entity which can b calculated (ex: bridge, tunnel) we have to add length of d train+length of entity,

Hence (x+264)/y = 20 ->2.

By solving 1 and 2 you will get answer.

Last one is very smart and short method. Actually I was also doing this. But I forgot to convert m/s to km/hr. For that my answer was 22 m/s. So I thought that I was wrong. Because the answer given in km/hr.

Total distance = 284.

And time taken (20-8) = 12 because it is in same direction.

And now Speed = Distance/Time.

= 284/12.

= 22 m/s.

= 22*18/5 km/hr.

= 79.2 km/hr.

1) Train cross the pole in 8 sec.

Speed = Distance/Time.

Assume the train speed is x and length of train be y.

x = y/8 or y = 8x.

2) Train cross the bride which is 264mts long 20 sec.

x = (264+y)/20.

20x = 264+8x because y = 8x in above.

Solving the above equation we get speed is 22m/sec.

Finally convert the m/sec to km/hr.

22*18/5 = 79.2 km/hr.

S=D/T.

GIVEN DATA IS D=264, T1=8, T2=20.
LET TRAIN DISTANCE IS X,& TRAIN SPEED IS Y.
SO,
Y=X/8.
X=8Y.

THEN RELATIVE SPEED = X+264/20.
here speed is Y.
Y=X+264/20.
we know X value.
Y=(8Y+264)/20,
20Y=8Y+264,
12Y=264,
Y=264/12= 22 m/s.
Now its convert into km/h.
22*18/5=396/5=79.2 km/h.

T = 8 Sec.
T+264 = 20 Sec.
---------------

T SPEED = 264/12 m/s = 22 m/s Where 12 Sec. = 20-8(Difference).
T = length of Train.
Speed = 22*18/5 = 79.2 km/hr Answer.
If calculate length of Train then 8 Sec*22 m/s = 176 m.

Relative time = 20 - 8
= 12
Then,

Speed = Distance/ time
= 264/12
= 22 m/sec
= 22 *15/5 km/h.
= 79.2 km/ h.

Answer = 79.2 km/ h.

****Simple technique************

let's length of train = x.
Sp. of train = x/8.
Mathematically, (x+264)/20 = x/8.
or, x=176 m.

Now, speed of train = (176/8)*(18/5) = 79.2 km/hr.