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?
Answer: Option
Explanation:
Let the length of the train be x metres and its speed by y m/sec.
Then, | x | = 8 ![]() |
y |
Now, | x + 264 | = y |
20 |
8y + 264 = 20y
y = 22.
![]() |
![]() |
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.
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.
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.
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.
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.
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.
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.
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.
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.
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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