Question

Q. A bus is moving at the speed of 40 2/5 km/h. How far will it travel in 6 2/10 hours? If the bus is still at a distance of 10 2/3 km from the final destination, how far is the final destination from the starting point?

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given}:\mathrm{Speed}\mathrm{of}\mathrm{car}=40\frac{2}{5}=\frac{202}{5}\mathrm{km}/\mathrm{hr} \\ \mathrm{Time}\mathrm{taken}\mathrm{in}\mathrm{travel}=6\frac{2}{10}=\frac{62}{10}\mathrm{hr}=\frac{31}{5}\mathrm{hr} \\ \mathrm{We}\mathrm{know}:\mathrm{Distance}=\mathrm{Speed}\times \mathrm{Time} \\ \mathrm{So},\mathrm{Total}\mathrm{distance}\mathrm{covered}\mathrm{by}\mathrm{car}=\frac{202}{5}\times \frac{31}{5}=\frac{6262}{25}=250\frac{12}{25}\mathrm{km} \\ \mathrm{Also}\mathrm{given}:\mathrm{Car}\mathrm{is}\mathrm{still}10\frac{2}{3}\mathrm{km}\mathrm{from}\mathrm{final}\mathrm{destination},\mathrm{So} \\ \mathrm{Distance}\mathrm{of}\mathrm{final}\mathrm{destination}=250\frac{12}{25}+10\frac{2}{3} \\ =\frac{6262}{25}+\frac{32}{3} \\ =\frac{18786+800}{75} \\ =\frac{19586}{75}\mathrm{km} \end{gathered}$$
Regards