At its usual speed, a train of length L metres crosses platform 300 metre long in 25 seconds. At 50% of its usual speed, the train crosses a vertical pole in 20 seconds. What is the value of L?

Let usual speed of the train = $$10x$$ m/s

Now, 50% of the speed = $$\frac{50}{100} \times 10x = 5x$$ m/s

Length of train = $$l$$ m

Time taken to cross the pole = 20 sec

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

=> $$5x = \frac{l}{20}$$

=> $$x = \frac{l}{100}$$

Length of platform = 300 m

Acc. to ques, => $$10x = \frac{300 + l}{25}$$

=> $$10 \times \frac{l}{100} = \frac{300 + l}{25}$$

=> $$\frac{l}{10} = \frac{300 + l}{25}$$

=> $$25l = 3000 + 10l$$

=> $$25l - 10l = 15l = 3000$$

=> $$l = \frac{3000}{15} = 200$$ m