Solution:
$$\eqalign{
& {\text{A's}}{\kern 1pt} {\kern 1pt} {\text{speed}} = \left( {5 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{25}}{{18}}{\text{m/sec}} \cr
& {\text{Time}}{\kern 1pt} {\kern 1pt} {\text{taken}}{\kern 1pt} {\kern 1pt} {\text{by}}{\kern 1pt} {\kern 1pt} {\text{A}}{\kern 1pt} {\kern 1pt} {\text{to}}{\kern 1pt} {\kern 1pt} {\text{cover 100 m}} \cr
& = \left( {100 \times \frac{{18}}{{25}}} \right)\sec \cr
& = 72\sec \cr
& \therefore {\text{Time}}{\kern 1pt} {\kern 1pt} {\text{taken}}{\kern 1pt} {\kern 1pt} {\text{by}}{\kern 1pt} {\kern 1pt} {\text{B}}{\kern 1pt} {\kern 1pt} {\text{to}}{\kern 1pt} {\kern 1pt} {\text{cover 92 m}} \cr
& = \left( {72 + 8} \right) = 80\sec \cr
& \therefore {\text{B's}}{\kern 1pt} {\kern 1pt} {\text{speed}} = \left( {\frac{{92}}{{80}} \times \frac{{18}}{5}} \right){\text{kmph}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.14{\text{ kmph}} \cr} $$
