In a race of 200 m, B can give a start of 10 m to A and C can give a start of 20 m to B. The start that C can give to A in the same race is

### Concept & Chained Start Margins When starts are given serially ($C \rightarrow B$ and $B \rightarrow A$), we convert these starts into distance ratios and multiply them to find the direct relationship between the first and last runner. ### Step-by-Step Solution - **Given:** Race = $200$ m. - $B$ gives $A$ a $10$ m start. When $B$ runs $200$ m, $A$ runs $190$ m. Ratio $\frac{B}{A} = \frac{200}{190} = \frac{20}{19}$. - $C$ gives $B$ a $20$ m start. When $C$ runs $200$ m, $B$ runs $180$ m. Ratio $\frac{C}{B} = \frac{200}{180} = \frac{10}{9}$. - To find the start $C$ can give $A$, we need the ratio $\frac{C}{A}$: $$ \frac{C}{A} = \left(\frac{C}{B}\right) \times \left(\frac{B}{A}\right) = \frac{10}{9} \times \frac{20}{19} = \frac{200}{171} $$ - This means when $C$ runs $200$ m, $A$ runs $171$ m. - The start $C$ can give $A$ is $200 - 171 = 29$ m. ### Exam Strategy & Shortcut Use fractional chaining directly on the total distance. If $C$ runs $200$m, $B$ runs $180$m. $B$ runs $\frac{180}{200}$ of the race. $A$ runs $\frac{190}{200}$ of whatever $B$ runs. So $A$ runs $\left(\frac{19}{20}\right) \times 180 = 171$m. The start is $200 - 171 = 29$m. ### Common Pitfall A common error is simple addition of the starts: $10 \text{ m} + 20 \text{ m} = 30 \text{ m}$. This ignores the fact that $A$'s speed is slower relative to $B$'s distance, and $B$ only covers $180$m when competing against $C$. ### Final Answer Therefore, the correct answer is **29 m**.