$A$ and $B$ take part in a $100 \text{ m}$ race. $A$ runs at $5 \text{ km an hour}$. $A$ gives $B$ a start of $8 \text{ m}$ and still beats him by $8 \text{ seconds}$. The speed of $B$ is

### Concept & Speed Conversions When a runner is given a start in both distance and time, we can calculate the exact time they spent running and the exact distance they covered to find their speed. $$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$ ### Step-by-Step Solution * **Convert A's speed:** $5 \text{ km/hr} = 5 \times \frac{5}{18} = \frac{25}{18} \text{ m/sec}$. * **Time taken by A:** To cover $100 \text{ m}$, $A$ takes $\frac{100}{\frac{25}{18}} = 100 \times \frac{18}{25} = 72 \text{ seconds}$. * $A$ beats $B$ by $8 \text{ seconds}$. This means $B$ takes $8 \text{ seconds}$ longer to finish his portion of the race. * **Time taken by B:** $72 + 8 = 80 \text{ seconds}$. * $A$ gave $B$ a start of $8 \text{ m}$, meaning $B$ only had to run $100 - 8 = 92 \text{ m}$. * **Speed of B in m/sec:** $\frac{92 \text{ m}}{80 \text{ s}} = \frac{23}{20} \text{ m/sec}$. * **Convert B's speed to km/hr:** $\frac{23}{20} \times \frac{18}{5} = \frac{23 \times 9}{50} = \frac{207}{50} = 4.14 \text{ km/hr}$. ### Exam Strategy & Shortcut Break it down logically: $A$ finishes in $72 \text{ s}$. $B$ takes $80 \text{ s}$ to run $92 \text{ m}$. You know $\frac{92}{80}$ is slightly greater than $1$. In km/hr, $\frac{18}{5} \times 1.15 = 4.14$. ### Common Pitfall Forgetting to adjust $B$'s distance for the $8 \text{ m}$ start, or subtracting the $8 \text{ seconds}$ from $A$'s time instead of adding it. If $A$ beats $B$, $B$ took more time. ### Final Answer Therefore, the correct answer is **4.14 km/hr**.