A train travelling at a speed of 60 km/hr crosses a platform in 20 sec

Case 1: Train crosses a platform in 20 seconds.
When a train crosses a platform, the total distance covered by the train is the sum of its own length and the length of the platform ($L_T + L_P$).
Distance = Speed $\times$ Time.
$L_T + L_P = V_T \times 20$.
$L_T + L_P = \frac{50}{3} \times 20 = \frac{1000}{3}$ meters. (Equation 1)

Case 2: Train crosses a person walking at 6 km/hr in the same direction in 12 seconds.
The speed of the person is 6 km/hr. Convert this to m/s.
$V_P = 6 \times \frac{5}{18} = \frac{30}{18} = \frac{5}{3}$ m/s.
When the train crosses a person moving in the same direction, we use the relative speed.
Relative speed = Speed of train – Speed of person (since they are in the same direction).
Relative speed $= V_T – V_P = \frac{50}{3} – \frac{5}{3} = \frac{45}{3} = 15$ m/s.
The distance covered by the train relative to the person is the length of the train ($L_T$).
Distance = Relative Speed $\times$ Time.
$L_T = 15 \times 12 = 180$ meters.

Now that we have the length of the train ($L_T = 180$ m), we can use Equation 1 to find the length of the platform ($L_P$).
$L_T + L_P = \frac{1000}{3}$
$180 + L_P = \frac{1000}{3}$
$L_P = \frac{1000}{3} – 180 = \frac{1000 – 180 \times 3}{3} = \frac{1000 – 540}{3} = \frac{460}{3}$ meters.
Convert $\frac{460}{3}$ to a decimal: $460 \div 3 \approx 153.33$.
So, $L_T = 180$ m and $L_P \approx 153.33$ m.
The question asks for the length of the train and that of the platform, respectively.