A train takes 5 minutes to cross a telegraphic post. Then time taken by anor train whose length is just double of first train and moving with same speed to cross a platform of its own length is :

Let the length of the train be x metres.
Time taken to cover x meters = 5 min
= (5 × 60) sec
= 300 sec
Speed of the train = $$\frac{{\text{x}}}{{300}}$$ m/sec
Length of the second train = 2x meters
Length of the platform = 2x meters
∴ Required time
$$\eqalign{ & = \left[ {\frac{{2{\text{x}} + 2{\text{x}}}}{{\left( {\frac{{\text{x}}}{{300}}} \right)}}} \right]{\text{sec}} \cr & = \left( {\frac{{4{\text{x}} \times 300}}{{\text{x}}}} \right){\text{sec}} \cr & = 1200\,{\text{sec}} \cr & = \frac{{1200}}{{60}}\,{\text{min}} \cr & = 20\,{\text{minutes}} \cr} $$