In racing over a distance d at uniform speed, A can beat B by 20 metres, B can beat C by 10 metres, and A can beat C by 28 metres. Then d, in metres, is

### Concept & Transitive Ratios in Distance When comparing the distances covered by racers in the same timeframe, the ratio of their distances remains constant regardless of the total race length. $$\frac{D_A}{D_C} = \left(\frac{D_A}{D_B}\right) \times \left(\frac{D_B}{D_C}\right)$$ ### Step-by-Step Solution * Let the total distance of the race be $d$. * When $A$ covers $d$ metres, $B$ covers $(d - 20)$ metres. Ratio $\frac{A}{B} = \frac{d}{d - 20}$. * When $B$ covers $d$ metres, $C$ covers $(d - 10)$ metres. Ratio $\frac{B}{C} = \frac{d}{d - 10}$. * When $A$ covers $d$ metres, $C$ covers $(d - 28)$ metres. Ratio $\frac{A}{C} = \frac{d}{d - 28}$. * Using the transitive property of ratios: $\frac{A}{C} = \frac{A}{B} \times \frac{B}{C}$. * Substitute the expressions: $\frac{d}{d - 28} = \left(\frac{d}{d - 20}\right) \times \left(\frac{d}{d - 10}\right)$. * Cancel one $d$ from the numerators on both sides: $\frac{1}{d - 28} = \frac{d}{(d - 20)(d - 10)}$. * Cross-multiply: $(d - 20)(d - 10) = d(d - 28)$. * Expand both sides: $d^2 - 30d + 200 = d^2 - 28d$. * Cancel $d^2$ from both sides: $-30d + 200 = -28d$. * $200 = 30d - 28d$ * $2d = 200$ * $d = 100$ metres. ### Exam Strategy & Shortcut Instead of full algebra, test the options. If $d = 100$: $\frac{A}{B} = \frac{100}{80} = \frac{5}{4}$. $\frac{B}{C} = \frac{100}{90} = \frac{10}{9}$. Then $\frac{A}{C} = \frac{5}{4} \times \frac{10}{9} = \frac{50}{36} = \frac{100}{72}$. If $A$ runs 100, $C$ runs 72. $A$ beats $C$ by $100 - 72 = 28$ metres. This perfectly matches the prompt! ### Common Pitfall A common mistake is assuming that beat distances simply add up (i.e., thinking $A$ beats $C$ by $20 + 10 = 30$ metres). Distances must be multiplied as ratios because they represent speeds. ### Final Answer Therefore, the correct answer is **100**.