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Let the length of the train be meters.

Case 1: when train passes a pole

Speed of the train,$s = \dfrac{{distance\,(train\,\,length)}}{{time}}$

$ \Rightarrow s = \dfrac{x}{5}m/s$ \[........\left( 1 \right)\]

Case 2: When train crosses the platform,

Speed of the train $ = \dfrac{{distance\,(platform\,\,length)}}{{time}}$

Since, the distance is length of the train and length of the platform (because train will pass the platform with its length),

Speed of train $ = $\[\dfrac{{length{\text{ }}of{\text{ }}train{\text{ }} + {\text{ length of }}platform}}{{Time}}\]

$ \Rightarrow s = \dfrac{{x + 50}}{{10}}m/s$ \[.......\left( 2 \right)\]

From \[\left( 1 \right){\text{ }}and{\text{ }}\left( 2 \right),\]we will get

$\dfrac{x}{5} = \dfrac{{x + 50}}{{10}}$

$ \Rightarrow \dfrac{x}{1} = \dfrac{{x + 50}}{2}$

By cross multiplying the terms in the above expression, we will get

\[2x = x + 50\]

\[ \Rightarrow 2x - {\text{x}} = 50\]

$ \Rightarrow x = 50$

$ \Rightarrow $Length of train\[ = 50\]meters.

$\left( i \right){\text{ }}x{\text{ }}km/hr = \left( {x \times \dfrac{5}{{18}}} \right)m/s.$

$\left( {ii} \right)\;x{\text{ }}m/s = \left( {x \times \dfrac{{18}}{5}} \right)km/hr.$