A train running at 48 km/h crosses a man going with the speed of 12 km/h, in the same direction, in 18 seconds and passes a woman coming from the opposite direction in 12 seconds. The speed (in km/h) of the woman is :

Let the length of the train = L

Relative speed between train and man = 48 - 12 = 36 km/h

= 36 x $$\frac{5}{18}$$ m/sec

= 10 m/sec

Time taken by train to cross the man = 18 seconds

$$\frac{L}{10}$$ = 18

L = 180 m

Length of the train = 180 m

Let the speed of the woman = s km/h

Relative speed between train and woman = (48 + s) km/h

= (48 + s) x $$\frac{5}{18}$$ m/sec

Time taken by train to cross the woman = 12 seconds

$$\frac{L}{\left(48+s\right)\times\frac{5}{18}}=12$$

$$\frac{180}{\left(48+s\right)\times\frac{5}{18}}=12$$

$$\frac{180\times18}{\left(48+s\right)5}=12$$

48 + s = 54

s = 6

Speed of the woman = 6 km/h

Hence, the correct answer is Option A