Aptitude - Problems on Trains - Discussion

Discussion Forum : Problems on Trains - General Questions (Q.No. 20)
20.
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?
69.5 km/hr
70 km/hr
79 km/hr
79.2 km/hr
Answer: Option
Explanation:

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 8     =>     x = 8y
y

Now, x + 264 = y
20

=> 8y + 264 = 20y

=> y = 22.

Therefore Speed = 22 m/sec = ( 22 x 18 ( km/hr = 79.2 km/hr.
5

Discussion:
46 comments Page 1 of 5.

Vishnu said:   3 years ago
D = 264m,
T1 = 8 s,
T2 = 20s,
T = 20-8,
T = 12sec,

S = D/T.
S= 264/12,
S= 22 m/s,
S * 5/18 = 22.
S = 22 * 18/5.
S = 79.2 km/hr.
(24)

PRIYADARSHI PRATTYULYA KUMAR BHATTA said:   4 years ago
Simply the solution is;

12sec = 264
1sec = 264/12 = 22.
i.e. speed is 22 m / sec.
So we convert it into km/hr.
That is 22*18/5 = 79.2 km/hr.
(19)

Ravibhai said:   2 years ago
Simple:

Time = Distance/Speed.
Speed = Distance/Time.
= 264/(20-8)×5/18.
= 79.2km/h.
(18)

Debashis said:   6 years ago
****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.
(7)

Souvanik saha said:   9 years ago
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.
(3)

Govindraj said:   6 years ago
Hi all,

22 m/s = 22*1000/3600 km/hr = 6.1 km/hr.

Let me know if I am wrong.
(2)

Babu Tunk said:   9 years ago
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.
(1)

Nagarjun said:   1 decade ago
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.
(1)

Vignesh said:   10 years ago
Don't go complex one, we need Speed = Distance/Time.

Total distance = 284.

Time taken to cross is 12 sec (20-8).

Speed = 22 m/s or 79.2 km/hr.
(1)

Aishuy said:   7 years ago
How will make 22*18/5? Please explain this step.


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