At its usual speed, a 150 metre long train crosses a platform of length L metres in 24 seconds. AT 75% of its usual speed, the train crosses a vertical pole in 12 seconds. What is the value of L?

Let speed of the train = $$20x$$ m/s

Now, 75% of the speed = $$\frac{75}{100} \times 20x = 15x$$ m/s

Length of train = 150 m

Time taken to cross the pole = 12 sec

Using, $$speed = \frac{distance}{time}$$

=> $$15x = \frac{150}{12}$$

=> $$x = \frac{10}{12} = \frac{5}{6}$$

Length of platform = $$l$$ m

Acc. to ques, => $$20x = \frac{150 + l}{24}$$

=> $$20 \times \frac{5}{6} = \frac{150 + l}{24}$$

=> $$\frac{50}{3} = \frac{150 + l}{24}$$

=> $$150 + l = 400$$

=> $$l = 400 - 150 = 250$$ m