A car travelling with $$\frac{5}{7}$$ of its usual speed covers 42 km in 1 hr 40 min 48 sec. What is usual speed of car ?

$$\eqalign{ & {\text{Usual speed}}\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,{\text{New speed}} \cr & \,\,\,\mathop {\,\,\,\,\,\,\, \downarrow \times 5}\limits_{35{\text{ km/h}}}^7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\mathop {\,\,\,\,\,\,\, \downarrow \times 5}\limits_{{\text{25 km/h}}}^5 \cr} $$
$$\because $$ Train covers 42 kms in 1 hr, 40 min, 48 sec with the speed of $$\frac{5}{7}$$ of its speed.
Then its new speed :
$$ = \frac{{{\text{Distance }}}}{{{\text{Time}}}} = \frac{{42{\text{ km}}}}{{\frac{{504}}{{300}}{\text{hr}}}}$$

\[\left\{ \begin{gathered} {\text{1 hr 40 min 48 sec}} \hfill \\ {\text{1hr + 40 min + }}\frac{{48}}{{60}}{\text{min}} \hfill \\ {\text{1 hr + }}\left( {40 + \frac{4}{5}} \right)\min \hfill \\ 1{\text{ hr}} + \frac{{204}}{5}\min \hfill \\ \left( {1 + \frac{{204}}{{5 \times 60}}} \right){\text{ hr}} = \frac{{504}}{{300}}{\text{ hr }} \hfill \\ \end{gathered} \right\}\]

$$\eqalign{ & = \frac{{42}}{{504}} \times 300{\text{ km/hr}} \cr & {\text{ = 25 km/hr}} \cr & \because 5{\text{ units = 25 km/hr}} \cr & \,\,\,\,{\text{1 unit = 5 km/hr}} \cr & \because {\text{Usual speed = 7 units}} \cr & \because {\text{Usual speed = 7}} \times {\text{5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,= 35{\text{ km/hr}} \cr} $$