A car travelling with 5/7 of its usual speed covers 42 km in 1 hr 40 min 48 sec. What is the usual speed of the car ?

Answer: Option B

$$\begin{array}{rl}& \text{Usual speed}:\text{New speed}\\ & \underset{35\text{ km/h}}{\overset{7}{\downarrow \times 5}}:\underset{\text{25 km/h}}{\overset{5}{\downarrow \times 5}}\end{array}$$
$$\because$$ 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 :
$$=\frac{\text{Distance }}{\text{Time}}=\frac{42\text{ km}}{\frac{504}{300}\text{hr}}$$

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

$$\begin{array}{rl}& =\frac{42}{504}\times 300\text{ km/hr}\\ & \text{ = 25 km/hr}\\ & \because 5\text{ units = 25 km/hr}\\ & \text{1 unit = 5 km/hr}\\ & \because \text{Usual speed = 7 units}\\ & \because \text{Usual speed = 7}\times \text{5}\\ & =35\text{ km/hr}\end{array}$$