Solution:
Due to stoppages, it covers 9 k less.
Time taken to cover 9 km :
$$\eqalign{ = \left( {\frac{9}{{54}} \times 60} \right){\text{ min}} \cr = 10\min \cr} $$

Solution:
Let the actual distance travelled be x km
Then,
$$\eqalign{ \Leftrightarrow \frac{x}{{10}} = \frac{{x + 20}}{{14}} \cr \Leftrightarrow 14x = 10x + 200 \cr \Leftrightarrow 4x = 200 \cr \Leftrightarrow x = 50{\text{ km}} \cr} $$

Solution:
Average speed :
$$\eqalign{ = \frac{{2 \times 60 \times 45}}{{60 + 45}} \cr = \frac{{1080}}{{21}}{\text{ km/hr}} \cr} $$
Time :
$$\eqalign{ = 5 + \frac{1}{4}{\text{ hr}} \cr {\text{ = }}\frac{{21}}{4}{\text{ hr}} \cr} $$
Distance travelled :
$$\eqalign{ {\text{ = }}\frac{{1080}}{{21}} \times \frac{{21}}{4} \cr = 270{\text{ km}} \cr} $$

Solution:
$$\eqalign{ {\text{Let}}\,{\text{distance}} = x\,km\,{\text{and}}\,{\text{usual}}\,{\text{rate}} = y\,kmph \cr {\text{Then}},\,\frac{x}{y} - \frac{x}{{y + 3}} = \frac{{40}}{{60}} \cr \Rightarrow 2y\left( {y + 3} \right) = 9x\,..........\left( i \right) \cr {\text{and}},\,\frac{x}{{y - 2}} - \frac{x}{y} = \frac{{40}}{{60}} \cr \Rightarrow y(y - 2) = 3x\,.............\left( {ii} \right) \cr {\text{On}}\,{\text{dividing}}\,\left( {\text{i}} \right)\,{\text{by}}\,\left( {{\text{ii}}} \right){\text{,}} \cr {\text{We}}\,{\text{get}}\,x = 40\,\text{km} \cr} $$

Solution:
Speed :
$$\eqalign{ = \left( {\frac{{600}}{{5 \times 60}}} \right){\text{ m/sec}} \cr = {\text{ 2 m/sec}} \cr = \left( {2 \times \frac{{18}}{5}} \right){\text{ km/hr}} \cr = {\text{ 7}}{\text{.2 km/hr}} \cr} $$

Solution:
$$\eqalign{ {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr = \frac{{162}}{{15 \times \frac{{18}}{5}}}{\text{hr}} \cr = \frac{{162 \times 5}}{{15 \times 18}}{\text{hr}} \cr = 3{\text{ hrs}} \cr} $$

Solution:
$$\eqalign{ {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr {\text{Time}} = \frac{{300 + 200}}{{25}} \cr {\text{Time}} = 20\sec \cr} $$