A runs 1²/₃ times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

Explanation:-

$$\eqalign{ & {\text{Ratio}}{\kern 1pt} {\text{of}}{\kern 1pt} {\text{speeds}}{\kern 1pt} {\text{of}}{\kern 1pt} {\text{A}}{\kern 1pt} {\text{and}}{\kern 1pt} {\text{B}} = \frac{5}{3}:1 = 5:3 \cr & {\text{Thus,}}\,{\text{in}}\,{\text{race}}\,{\text{of}}\,{\text{5m,}}\,{\text{A}}\,{\text{gains}}\,{\text{2m}}\,{\text{over}}\,{\text{B}} \cr & {\text{2m}}\,{\text{are}}\,{\text{gained}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{a}}\,{\text{race}}\,{\text{of}}\,{\text{5m}} \cr & {\text{80m}}\,{\text{will}}\,{\text{be}}\,{\text{gained}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{a}}\,{\text{race}}\,{\text{of}}\, \cr & \left( {\frac{5}{2} \times 80} \right)m = 200m \cr & \therefore {\text{Winning}}\,{\text{post}}\,{\text{is}}\,{\text{200m}}\,{\text{away}}\,{\text{from}}\,{\text{the}}\,{\text{starting}}\,{\text{point}} \cr}$$