Answer: Option C
Suppose A covers 400 m in t seconds
Then, B covers 385 m in (t + 5) seconds
$$\eqalign{
& \therefore {\text{ B covers 400 m in}} \cr
& = \left\{ {\frac{{\left( {t + 5} \right)}}{{385}} \times 400} \right\}{\text{ sec}} \cr
& = \frac{{80\left( {t + 5} \right)}}{{77}}{\text{ sec}} \cr
& {\text{Also, B covers 400 m in}} \cr
& = \left( {t + 7\frac{1}{7}} \right){\text{ sec}} \cr
& = \frac{{\left( {7t + 50} \right)}}{7}{\text{ sec}} \cr
& \therefore \frac{{80\left( {t + 5} \right)}}{{77}} = \frac{{7t + 50}}{7} \cr
& \Rightarrow 80\left( {t + 5} \right) = 11\left( {7t + 50} \right) \cr
& \Rightarrow \left( {80t - 77t} \right) = \left( {550 - 400} \right) \cr
& \Rightarrow 3t = 150 \cr
& \Rightarrow t = 50 \cr
& \therefore {\text{ A's speed}} \cr
& = \frac{{400}}{{50}}\,m/\sec \cr
& = 8\,m/\sec \cr
& \therefore {\text{B's speed}} \cr
& = \frac{{385}}{{55}}\,m/\sec \cr
& = 7\,m/\sec \cr} $$