Aptitude - Problems on Trains - Discussion

Sp of man = 5 * 5/18
= 25/18 m/sec.

Let Sp of train be x, then;
x - 25/18 = 125/10
So x = 125/10 + 25/18 = 125/9 m/sec.
So x = 125/9 * 18/5 =50 Km/h.

l = 125m,
u = 5km/hr,
v = ?
t = 10s.
u - v = 125/10 m/s,
u - v = 25/2 m/s,
u - v = 5*9 km/hr,
u - v = 45km/hr,
5km/hr - v = 45km/hr.
v = 50 km/hr.
Hence, option (B).

Given
Length of train = 125m
Speed of man = 5km/hr = 5 * 5/18 = 25/18
Time = 10sec.
Speed = distance/time
Man running in the same direction so;

Let theh speed of the train is x.
speed = x - 25/18;
then by formula;

x - 25/18 = 125/10
x = 125/10 + 25 /18
x = 250/18 sec/m

Then;
convert into km/hr.
speed = 250/18 * 18/5.
speed = 50 => ans.

According to me, it should be 25/2 × 18/5 = 45.

@All.

Why do we minus the speed of the man moving in the same direction? Please explain to me.

The speed of train is 45 km per hour
And relative speed is 50 km per hour.
But in the question only speed of train is asked not the relative speed.
So, please either change the question and mention relative speed or change the answer to option A 45 km per hour which is the speed of train.

How it comes 18/5? Anyone, explain to me.

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.

S = L/T.
S = 125/10
S = 25/2.

then,
M/sec to km/hr convert.
1km/hr = [1000m/3600sec] = 5/18m/sec,
m/sec = 18/5km/hr.

i.e
(25/2 * 18/5)km/hr,
45km/hr.

Therefore first train x-5 = 45.
X = 50 km/hr.

Why was the formula reversed from 5/18 to 18/5? Please explain me.