Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 2)
2.
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
Answer: Option
Explanation:
Speed of the train relative to man = | ![]() |
125 | ![]() |
10 |
= | ![]() |
25 | ![]() |
2 |
= | ![]() |
25 | x | 18 | ![]() |
2 | 5 |
= 45 km/hr.
Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
x - 5 = 45
x = 50 km/hr.
Discussion:
475 comments Page 47 of 48.
Charanjeet pal said:
5 years ago
Why divide 18/5? Please explain.
Sai krishna. B said:
5 years ago
Why 60*60/1000 came here? Can you please explain?
Dinesh kumar said:
5 years ago
The speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
Please explain it, friends.
Please explain it, friends.
Wanbok Rabon said:
4 years ago
How get 18/5?
Yash said:
4 years ago
Why they have done 125/10?
Chemayek said:
4 years ago
Good explanation. Thanks all.
Tushar Patil said:
4 years ago
First we write Data:
We Have D=125m t=10s,
S=D/t =125/10=25/2 m/s.
We want to convert from m/s to km/kr.
so we multiply 25/2 into 18/5,
@When we convert km/hr to m/s we multiply by 5/18,
@When we have to convert m/s to km/hr we multiply by 18/5,
Our answer are given in km/hr so we convert according_to by multiplying 18/5.
=25/2*18/5=45 Km/hr,
let we assume train speed be X.
And Man 5 km/hr.
If given in opposite direct then we add,
Here have in the same direction so we subtract,
x-5=45 So Answer is 50.
We Have D=125m t=10s,
S=D/t =125/10=25/2 m/s.
We want to convert from m/s to km/kr.
so we multiply 25/2 into 18/5,
@When we convert km/hr to m/s we multiply by 5/18,
@When we have to convert m/s to km/hr we multiply by 18/5,
Our answer are given in km/hr so we convert according_to by multiplying 18/5.
=25/2*18/5=45 Km/hr,
let we assume train speed be X.
And Man 5 km/hr.
If given in opposite direct then we add,
Here have in the same direction so we subtract,
x-5=45 So Answer is 50.
Rohan said:
4 years ago
No, it's wrong!
At last, speed came 45km. Now since it's in the same direction hence we will subtract, right.
So S=S (train (since its bigger value) - the speed given (5) ) ==>45-5 = 40.
Speed cannot be negative hence we always subtract smaller speed value FROM bigger speed value in the case for the same direction!
At last, speed came 45km. Now since it's in the same direction hence we will subtract, right.
So S=S (train (since its bigger value) - the speed given (5) ) ==>45-5 = 40.
Speed cannot be negative hence we always subtract smaller speed value FROM bigger speed value in the case for the same direction!
Qasim said:
4 years ago
The train and the man are moving in the same direction we have to use subtraction.
Here, we have to find the value of X. So for that, we have to keep X separate either left side or right.
Shift 45 to left and x to right. Because we are shifting, the sign will be change (positive to negative & negative to positive).
x-5 = 45
-5 - 45 = -x
-50 = -x ....(- - = + , so - - canceled)
x = 50.
Here, we have to find the value of X. So for that, we have to keep X separate either left side or right.
Shift 45 to left and x to right. Because we are shifting, the sign will be change (positive to negative & negative to positive).
x-5 = 45
-5 - 45 = -x
-50 = -x ....(- - = + , so - - canceled)
x = 50.
Namith said:
4 years ago
What is the difference between the relative speed of train and the speed of the train? Anyone, please explain.
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