A runs $1 \frac{2}{3}$ times as fast as $B$. If $A$ gives $B$ a start of $80 \text{ m}$, how far must the winning post be so that $A$ and $B$ might reach it at the same time?

### Concept & Proportional Distances When two runners complete a race in the same time, the ratio of the distances they cover is equal to the ratio of their speeds. $$\frac{D_1}{D_2} = \frac{S_1}{S_2}$$ ### Step-by-Step Solution * **Given:** The speed of $A$ is $1 \frac{2}{3}$ times the speed of $B$. This means the ratio of their speeds is $S_A : S_B = \frac{5}{3} : 1 = 5 : 3$. * $A$ gives $B$ a start of $80 \text{ m}$. Let the winning post be at a distance of $x \text{ m}$. * To finish at the same time, $A$ must cover the full distance $x \text{ m}$, while $B$ must cover $(x - 80) \text{ m}$. * Since times are equal, $\frac{x}{x - 80} = \frac{5}{3}$. * Cross-multiplying gives: $3x = 5(x - 80)$. * $3x = 5x - 400$ * $2x = 400$ * $x = 200 \text{ m}$. ### Exam Strategy & Shortcut Use ratios directly. The speed ratio is $5:3$. This means in the same time, $A$ covers $5 \text{ units}$ of distance while $B$ covers $3 \text{ units}$. The difference is $2 \text{ units}$. We are given this difference is the $80 \text{ m}$ start. If $2 \text{ units} = 80 \text{ m}$, then $1 \text{ unit} = 40 \text{ m}$. $A$'s total distance is $5 \text{ units}$, so $5 \times 40 = 200 \text{ m}$. ### Common Pitfall Misinterpreting the fraction $1 \frac{2}{3}$ or incorrectly setting up the distance for $B$ as $x + 80$ instead of $x - 80$. ### Final Answer Therefore, the correct answer is **200 m**.