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 4 of 5.

Rana sk said:   8 years ago
20-8=12,
264÷12=22
22*18÷5=79.2.

Ginil kumar said:   7 years ago
@Rana Sk.

This is the simple way to get this answer. Thanks for explaining.

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.

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.

Sainpal said:   7 years ago
Nice method. Thanks @Diksha Rajput.

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?

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

Shambhu said:   7 years ago
Hi, I can't understand please explain me.

Laxman biradar said:   7 years ago
Nice explanation @Gaurav Reddy.

Prashant said:   6 years ago
Thanks @Randhir.


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