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.
Priya said:
1 decade ago
Hi, I didn't understand the solution. Can someone explain step by step ?
Randhir said:
1 decade ago
@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
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
Mani said:
1 decade ago
Please any one explain why we divide 176/8?
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)
Gai3 said:
1 decade ago
Thanks @Nagarjun.
Sanchita said:
1 decade ago
Please explain sum one in simple method.
Angel said:
1 decade ago
Hey can anyone tell me instead of taking x/y=8 , Can we take x/y=20 because it is also time ?
Janaka said:
1 decade ago
In question No. 19 x/y=15 is mentioned as x/15=y. But in this question when x/y=8 is mentiond as x=8y. What is the reason for this? please explain me.
Pavan said:
1 decade ago
Why we didn't add time. While calculating the speed. We taken only 20. Why we didn't take 20+8?
PAKA RAKESH said:
1 decade ago
Simply take time(20-8)=12 secs
speed=264*(18/5)=79.2km/hr
speed=264*(18/5)=79.2km/hr
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