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:
487 comments Page 2 of 49.
Sonianenavath said:
7 months ago
Very Good Thanks.
(7)
Sanika prakash said:
8 months ago
Good answer. Thanks all for explaining the answer.
(14)
Aathisha Sadhasivam said:
9 months ago
How it 18÷5 will become? Please explain.
Here is the solution for this question:
The length of a train is given as 125m (here it is mentioned as m/s), and the speed is given as 5km/hr (here it is mentioned as km/hr).
While converting km/hr into m/s, we are multiplying the speed by 5/18; therefore while converting m/s into km/hr, we have to take the reciprocal of 5/18, that is 18/5. The speed of the train should be calculated in km/hr; therefore, we take 18/5.
Here is the solution for this question:
The length of a train is given as 125m (here it is mentioned as m/s), and the speed is given as 5km/hr (here it is mentioned as km/hr).
While converting km/hr into m/s, we are multiplying the speed by 5/18; therefore while converting m/s into km/hr, we have to take the reciprocal of 5/18, that is 18/5. The speed of the train should be calculated in km/hr; therefore, we take 18/5.
(32)
Suja said:
9 months ago
How it 18÷5 will become? Please explain.
(16)
Preetilata Das said:
9 months ago
Someone, please explain this question properly.
(11)
Naveen reddy said:
10 months ago
125/10 = 12.5,
12.5 * 3.6 = 45,
45 + 5 = 50.
12.5 * 3.6 = 45,
45 + 5 = 50.
(35)
Anamuddin said:
1 year ago
Since the man is running in the same direction as the train, the distance covered by the train relative to the man is indeed the sum of the distance travelled by the man and the length of the train.
Let's recalculate:
Step 1: Calculate the distance travelled by the man in 10 seconds
Distance traveled by man (d_m) = Speed of man × Time = 1.39 m/s × 10 s = 13.9 m.
Step 2: Calculate the total distance covered by the train relative to the man
Total distance (d_t) = Length of train + Distance traveled by man = 125 m + 13.9 m = 138.9 m.
Step 3: Calculate the relative speed of the train concerning the man
Relative speed (v_r) = Total distance / Time = 138.9 m/10 s = 13.89 m/s
Step 4: Convert the relative speed to kilometres per hour.
Relative speed (v_r) = 13.89 m/s × (3600s/1000 m) = 50 km/hr.
Step 5: Calculate the speed of the train.
Speed of the train (v_t) = Relative speed + Speed of the man = 50 km/hr + 5 km/hr = 55 km/hr.
The correct answer is:
55 km/hr.
Let's recalculate:
Step 1: Calculate the distance travelled by the man in 10 seconds
Distance traveled by man (d_m) = Speed of man × Time = 1.39 m/s × 10 s = 13.9 m.
Step 2: Calculate the total distance covered by the train relative to the man
Total distance (d_t) = Length of train + Distance traveled by man = 125 m + 13.9 m = 138.9 m.
Step 3: Calculate the relative speed of the train concerning the man
Relative speed (v_r) = Total distance / Time = 138.9 m/10 s = 13.89 m/s
Step 4: Convert the relative speed to kilometres per hour.
Relative speed (v_r) = 13.89 m/s × (3600s/1000 m) = 50 km/hr.
Step 5: Calculate the speed of the train.
Speed of the train (v_t) = Relative speed + Speed of the man = 50 km/hr + 5 km/hr = 55 km/hr.
The correct answer is:
55 km/hr.
(41)
Phanikumar said:
1 year ago
How it comes 18/5? Anyone, explain to me.
(36)
Ajit Singh said:
1 year ago
Step 1:
The relative speed of the train concerning man = Distance/Time =125m/10sec= 12.5m/s.
Step 2: Convert the answer of "step 1" from m/s to km/hr..
To convert it into km/hr multiply it with 18/5.
12.5 * 18/5 = 45km/hr.
Step 3:
Train as well as the man both are moving in the same direction so we will have to subtract the speed of train(x) from the speed of the man(5).
(Note: If the train and man both are moving in opposite directions then add both of their values (x+5) ).
Relative speed = (x-5)km/hr
45km/hr = (x-5),
x=50 km/hr.
The relative speed of the train concerning man = Distance/Time =125m/10sec= 12.5m/s.
Step 2: Convert the answer of "step 1" from m/s to km/hr..
To convert it into km/hr multiply it with 18/5.
12.5 * 18/5 = 45km/hr.
Step 3:
Train as well as the man both are moving in the same direction so we will have to subtract the speed of train(x) from the speed of the man(5).
(Note: If the train and man both are moving in opposite directions then add both of their values (x+5) ).
Relative speed = (x-5)km/hr
45km/hr = (x-5),
x=50 km/hr.
(80)
Ameya said:
1 year ago
Given:
Length of the train = 125 m.
Speed of the man = 5 km/hr.
Time taken to pass the man = 10 seconds.
First, we need to convert the speed of the man from kilometres per hour to meters per second:
5 km hr=5×1000 m3600 s=50003600≈1.39 m s.
5 km hr=5×3600 s1000 m=36005000≈1.39 m s.
Next, we calculate the speed of the train using the formula:
Speed=DistanceTime.
Speed=TimeDistance.
The distance in this case is equal to the length of the train (125 m), and it passes the man in 10 seconds:
Speed of train relative to man =125 m10 s=12.5 m s.
Speed of train relative to man =10 s125 m=12.5 m s.
Since this is the relative speed (train's speed minus man's speed), we can express it as:
Speed of train − Speed of man = 12.5 m s.
Speed of train − Speed of man =12.5 m s.
Substituting in the man's speed:
Speed of train−1.39 m s=12.5 m s
Speed of train−1.39 m s=12.5 m s.
Now, solving for the speed of the train:
Speed of train=12.5+1.39=13.89 m s
Speed of train=12.5+1.39=13.89 m s
Finally, converting this back to kilometres per hour:
13.89 m s=13.89×36001000=50 km hr
13.89 m s=13.89×10003600=50 km hr.
Therefore, the speed of the train is 50 km/hr.
Length of the train = 125 m.
Speed of the man = 5 km/hr.
Time taken to pass the man = 10 seconds.
First, we need to convert the speed of the man from kilometres per hour to meters per second:
5 km hr=5×1000 m3600 s=50003600≈1.39 m s.
5 km hr=5×3600 s1000 m=36005000≈1.39 m s.
Next, we calculate the speed of the train using the formula:
Speed=DistanceTime.
Speed=TimeDistance.
The distance in this case is equal to the length of the train (125 m), and it passes the man in 10 seconds:
Speed of train relative to man =125 m10 s=12.5 m s.
Speed of train relative to man =10 s125 m=12.5 m s.
Since this is the relative speed (train's speed minus man's speed), we can express it as:
Speed of train − Speed of man = 12.5 m s.
Speed of train − Speed of man =12.5 m s.
Substituting in the man's speed:
Speed of train−1.39 m s=12.5 m s
Speed of train−1.39 m s=12.5 m s.
Now, solving for the speed of the train:
Speed of train=12.5+1.39=13.89 m s
Speed of train=12.5+1.39=13.89 m s
Finally, converting this back to kilometres per hour:
13.89 m s=13.89×36001000=50 km hr
13.89 m s=13.89×10003600=50 km hr.
Therefore, the speed of the train is 50 km/hr.
(28)
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