Let the length of the train be x m.

According to the question,

\( \Large \frac{x}{9}=Speed \) ...(i)

\( \Large \frac{x+150}{15}=Speed \) ...(ii)

From Eqs. (i) and (ii), we get

\( \Large \frac{x}{9} = \frac{x+150}{15} \)

=> \( \Large \frac{x}{3} = \frac{x+150}{5} \)

=> 5x = 3x + 450

=> x = 225 m