A car travelling with $$\frac{5}{7}$$ of its usual speed covers 42 km in 1 hr 40 min 48 sec. What is the usual speed of the car ?
A. $$17\frac{6}{7}$$ km/hr
B. 35 km/hr
C. 25 km/hr
D. 30 km/hr
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {\text{Usual speed}}\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,{\text{New speed}} \cr & \,\,\,\mathop {\,\,\,\,\,\,\, \downarrow \times 5}\limits_{35{\text{ km/h}}}^7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\mathop {\,\,\,\,\,\,\, \downarrow \times 5}\limits_{{\text{25 km/h}}}^5 \cr} $$$$\because $$ 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 :
$$ = \frac{{{\text{Distance }}}}{{{\text{Time}}}} = \frac{{42{\text{ km}}}}{{\frac{{504}}{{300}}{\text{hr}}}}$$
\[\left\{ \begin{gathered} {\text{1 hr 40 min 48 sec}} \hfill \\ {\text{1hr + 40 min + }}\frac{{48}}{{60}}{\text{min}} \hfill \\ {\text{1 hr + }}\left( {40 + \frac{4}{5}} \right)\min \hfill \\ 1{\text{ hr}} + \frac{{204}}{5}\min \hfill \\ \left( {1 + \frac{{204}}{{5 \times 60}}} \right){\text{ hr}} = \frac{{504}}{{300}}{\text{ hr }} \hfill \\ \end{gathered} \right\}\]
$$\eqalign{ & = \frac{{42}}{{504}} \times 300{\text{ km/hr}} \cr & {\text{ = 25 km/hr}} \cr & \because 5{\text{ units = 25 km/hr}} \cr & \,\,\,\,{\text{1 unit = 5 km/hr}} \cr & \because {\text{Usual speed = 7 units}} \cr & \because {\text{Usual speed = 7}} \times {\text{5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,= 35{\text{ km/hr}} \cr} $$
Related Questions on Speed Time and Distance
A. 48 min.
B. 60 min.
C. 42 min.
D. 62 min.
E. 66 min.
A. 262.4 km
B. 260 km
C. 283.33 km
D. 275 km
E. None of these
A. 4 hours
B. 4 hours 30 min.
C. 4 hours 45 min.
D. 5 hours
Join The Discussion