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  x = 8y |
| y |
| Now, | x + 264 | = y |
| 20 |
8y + 264 = 20y
y = 22.
 Speed = 22 m/sec = |
 |
22 x | 18 |  |
km/hr = 79.2 km/hr. |
| 5 |
Discussion:
46 comments Page 2 of 5.
Akhila said:
7 years ago
Please explain this.
After finding the relative speed 20-8=12secs, why we are not taking the total distance as x+264?
After finding the relative speed 20-8=12secs, why we are not taking the total distance as x+264?
Sainpal said:
7 years ago
Nice method. Thanks @Diksha Rajput.
Diksha rajput said:
7 years ago
It can be done directly.
D=264m T= (20-8) =12 s.
S=D/T =264/12 =22.
Converting it into kmph.
22*18/5=79.2.
D=264m T= (20-8) =12 s.
S=D/T =264/12 =22.
Converting it into kmph.
22*18/5=79.2.
Guravareddy namburi said:
7 years ago
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.
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.
Ginil kumar said:
7 years ago
@Rana Sk.
This is the simple way to get this answer. Thanks for explaining.
This is the simple way to get this answer. Thanks for explaining.
Rana sk said:
8 years ago
20-8=12,
264÷12=22
22*18÷5=79.2.
264÷12=22
22*18÷5=79.2.
Manu said:
8 years ago
We directly do speed=264/12.
=22*18/5. And we get the answer.
=22*18/5. And we get the answer.
Aarthy said:
8 years ago
@Hussaian.
To convert m/s to km/be, we have the formula a* 18/5 because to convert the metre per sec to km/h we multiply the number to b converted by 3600/1000 by simplifying this we get 18/5.
To convert m/s to km/be, we have the formula a* 18/5 because to convert the metre per sec to km/h we multiply the number to b converted by 3600/1000 by simplifying this we get 18/5.
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)
Raghavi said:
9 years ago
I am not getting the answer when we proceed it with the same step as the previous problem.
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