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:
478 comments Page 32 of 48.
Sankar said:
1 decade ago
Man has traveled 13.8 m in 10 sec.
i.e 5km/hr = 1.38m/s.
1.38*10s=13.8m he has traveled in 10 sec.
So 125m train has passed 13.8m in 10 sec.
So speed of the train is 13.8m/s i.e 49.68km/hr = 50 km/hr.
i.e 5km/hr = 1.38m/s.
1.38*10s=13.8m he has traveled in 10 sec.
So 125m train has passed 13.8m in 10 sec.
So speed of the train is 13.8m/s i.e 49.68km/hr = 50 km/hr.
Srinivas Rao said:
1 decade ago
If both things are in the same direction, then we have to do subtraction right but why we have to do it.
SURESHBABU KASTURI said:
1 decade ago
If two things are at same speed in same direction.
We subtract that,
So right 125-5=120.
Wrong is 125+5=130.
We subtract that,
So right 125-5=120.
Wrong is 125+5=130.
Selvakum said:
1 decade ago
(u-v). Using the given data find out the relative speed i.e. 45.
Vaishnapriya said:
1 decade ago
Consider x as the train speed(which is to be determined). We know a relative speed formula in same direction,
(u-v).
Using the given data find out the relative speed i.e 45.
Now,
Relative speed = (train speed-5).
(u-v).
Using the given data find out the relative speed i.e 45.
Now,
Relative speed = (train speed-5).
ESWARAN.P said:
1 decade ago
Here using x-5=45. Please explain?
Bikramjit said:
1 decade ago
1 km/hr -------------> 5/18 m/sec.
5/18 m/sec ------------> 1 km/hr.
1 m/sec -------------> 1/(5/18)km/hr = 18/5 km/hr.
25/2 m/sec -----------> (18/5)*(25/2) = 45 km/hr.
If train speed x km/hr, then (x-5)km/hr.
x-5 = 45.
x = 45+5.
x = 50 km/hr(answer).
5/18 m/sec ------------> 1 km/hr.
1 m/sec -------------> 1/(5/18)km/hr = 18/5 km/hr.
25/2 m/sec -----------> (18/5)*(25/2) = 45 km/hr.
If train speed x km/hr, then (x-5)km/hr.
x-5 = 45.
x = 45+5.
x = 50 km/hr(answer).
Eric said:
1 decade ago
@Shafi.
This method is quite simple and explain the whole concept, thanks for the nice explanation.
This method is quite simple and explain the whole concept, thanks for the nice explanation.
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.
Shanker said:
1 decade ago
We know that 1 km = 1000 meters,
1 hr = 60 minutes, 1 minute = 60 seconds.
And 1 hr = 60*60 seconds (i.e.360 seconds).
Then 1 km/hr = 1000 meters/360 seconds.
This becomes 5 meters/18 seconds((5/18) m/s).
So 1km/hr=5/18 m/s //line1.
But in this question it is given in m/s.
Then from the above line 1 we can write 1 m/s=18/5 km/hr.
1 hr = 60 minutes, 1 minute = 60 seconds.
And 1 hr = 60*60 seconds (i.e.360 seconds).
Then 1 km/hr = 1000 meters/360 seconds.
This becomes 5 meters/18 seconds((5/18) m/s).
So 1km/hr=5/18 m/s //line1.
But in this question it is given in m/s.
Then from the above line 1 we can write 1 m/s=18/5 km/hr.
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