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 |  m/sec |
| 10 |
| = | |
25 | m/sec. |
| 2 |
| = | |
25 | x | 18 | km/hr |
| 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 8 of 48.
K. Abhishek said:
6 years ago
@All.
Here, Train took 10 seconds to cross the person who is running at the speed of 5km/hr, then during this 10s the person might have travelled a certain distance, right? so train travelled (125m + distance covered by the man in 10s) in 10s then why the relative speed of train = 125/10?
How, will anyone explain it for me?
Here, Train took 10 seconds to cross the person who is running at the speed of 5km/hr, then during this 10s the person might have travelled a certain distance, right? so train travelled (125m + distance covered by the man in 10s) in 10s then why the relative speed of train = 125/10?
How, will anyone explain it for me?
Manishkumar said:
1 decade ago
@ all apply logic don't simply go for routine
speed of man is 5km/hr and in mt/sec is 1.39.
so total distance covered by main in 10 sec is 13.9 m
so total distance covered by train in 10 sec is 125+13.9=138.9
so total distance covered by train in 1 sec is 13.89m
and if we convert this in km/hr
i.e. 13.89*3600/1000= 50km/hr
speed of man is 5km/hr and in mt/sec is 1.39.
so total distance covered by main in 10 sec is 13.9 m
so total distance covered by train in 10 sec is 125+13.9=138.9
so total distance covered by train in 1 sec is 13.89m
and if we convert this in km/hr
i.e. 13.89*3600/1000= 50km/hr
Basava g said:
8 years ago
First, we find total length to across moving man=125+((5*10^3)÷(3600))*10.
= 125+13.888.
= 138.88m.
The speed of train = total/time.
= 138/10,
= 13.88m/s.
This one convert into km/hrs.
= 13.88 * 3600,
= 50km/hrs.
= 125+13.888.
= 138.88m.
The speed of train = total/time.
= 138/10,
= 13.88m/s.
This one convert into km/hrs.
= 13.88 * 3600,
= 50km/hrs.
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!
SHAFI said:
1 decade ago
Simple trick to handle:
S= D/T Where S-speed, D-distance, T-time
Now:
Train data.
D = 125m, T = 10sec.
find speed S = 125m/10sec = 25/2mps.
Note object running in same direction of train will be positive and in opposite direction will be negative.
SPEED OF MEN 5 km/hr+(25/2x5/18).
ANSWER 50 KM/HR.
S= D/T Where S-speed, D-distance, T-time
Now:
Train data.
D = 125m, T = 10sec.
find speed S = 125m/10sec = 25/2mps.
Note object running in same direction of train will be positive and in opposite direction will be negative.
SPEED OF MEN 5 km/hr+(25/2x5/18).
ANSWER 50 KM/HR.
Nani said:
9 years ago
Thank you friends, now I get clarity.
I am thinking speed is 45kmph and boy running speed is 5kmph so same direction we need to subtract the speed so I got 40kmph. But now I clarified we got 45kmph is already subtracted with 5kmph so now we need to find train speed so we must add 5kmph to 45kmph 50kmph is ams.
I am thinking speed is 45kmph and boy running speed is 5kmph so same direction we need to subtract the speed so I got 40kmph. But now I clarified we got 45kmph is already subtracted with 5kmph so now we need to find train speed so we must add 5kmph to 45kmph 50kmph is ams.
Hemanth Sairam said:
2 years ago
Let's think man is standing like a pole
125mtrs/10 secs = 12.5 mtrs per sec,
Convert seconds to hours;
12.5 mtrs x 60mins x 60 seconds = 45000 mtrs per hr,
= 45kmph.
The man also moving with some speed right, so the train will take some extra speed to cross than the man
So 45+5 = 50kmph actual train speed.
125mtrs/10 secs = 12.5 mtrs per sec,
Convert seconds to hours;
12.5 mtrs x 60mins x 60 seconds = 45000 mtrs per hr,
= 45kmph.
The man also moving with some speed right, so the train will take some extra speed to cross than the man
So 45+5 = 50kmph actual train speed.
(96)
Nirmal K Sahoo said:
1 decade ago
Man speed 5 km/hr, So man's speed per second (5*5/18) m/s.
After 10 sec, train passes the man, so, man cover distance in 10 sec is :
5*5*10/18 = 13.89m.
So, train covered total distance in 10 sec is :
125 m+13.89 m = 138.89 m.
So, speed of train is :138.89/10 m/s = 13.889*3600/1000 Km/hr = 50 Km/hr.
After 10 sec, train passes the man, so, man cover distance in 10 sec is :
5*5*10/18 = 13.89m.
So, train covered total distance in 10 sec is :
125 m+13.89 m = 138.89 m.
So, speed of train is :138.89/10 m/s = 13.889*3600/1000 Km/hr = 50 Km/hr.
Roopa said:
1 decade ago
Hi friends,
Here length of train = speed*time.
Likewise speed of train = length/time.
-> 125/10 = (25/2)m/sec.
Now given 5km/hr.
So speed is (x-5)km/hr or (x-5)(5/18).
Here we goes -> (25/2)m/sec = (x-5)(5/18).
-> (25/2)*(18/5) = (x-5).
-> 45 = (x-5).
-> x = 45+5.
-> x = 50 km/hr.
Here length of train = speed*time.
Likewise speed of train = length/time.
-> 125/10 = (25/2)m/sec.
Now given 5km/hr.
So speed is (x-5)km/hr or (x-5)(5/18).
Here we goes -> (25/2)m/sec = (x-5)(5/18).
-> (25/2)*(18/5) = (x-5).
-> 45 = (x-5).
-> x = 45+5.
-> x = 50 km/hr.
Pramod.M said:
9 years ago
Length(Distance) of the train 125m, time is 10 sec. The speed of man is 5 km/hr.
Speed = D/T i.e. 125/10 = 25/2m/s.
Therefore speed in km = 25/2 * 18/5 we got 45km/hr.
Moving in the same direction we must subtract i.e. 45 - 5 = 40 km/hr.
If the man moves against the train we must add 5km/hr to 45 km/hr.
Speed = D/T i.e. 125/10 = 25/2m/s.
Therefore speed in km = 25/2 * 18/5 we got 45km/hr.
Moving in the same direction we must subtract i.e. 45 - 5 = 40 km/hr.
If the man moves against the train we must add 5km/hr to 45 km/hr.
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