A man can reach a certain place in 30 hours. If he reduces his speed by $\frac{1}{15}$th, he goes 10 km less in that time. Find his speed. (S.S.C., 2002)

### Concept & Linear Distance Distance is directly proportional to speed when time is held constant. $$D \propto S$$ A reduction in speed translates directly to a proportional reduction in distance covered over the same duration. ### Step-by-Step Solution * Let the man's usual speed be $S$ km/hr. * The usual distance covered in 30 hours is $30S$. * He reduces his speed by $\frac{1}{15}$th, so his new speed is $S - \frac{1}{15}S = \frac{14}{15}S$. * The new distance covered in 30 hours is $30 \times \frac{14}{15}S = 28S$. * We are given that he goes 10 km less in this new scenario. * $30S - 28S = 10$ * $2S = 10 \Rightarrow S = 5$ km/hr. ### Exam Strategy & Shortcut Since time is constant (30 hours), the reduction in distance is solely due to the reduction in speed. Reduction in speed = $\frac{1}{15}S$. Distance lost = Reduction in Speed $\times$ Time = $(\frac{1}{15}S) \times 30 = 2S$. Given $2S = 10$, so $S = 5$. ### Common Pitfall A frequent error is calculating the new speed as $\frac{1}{15}S$ rather than subtracting $\frac{1}{15}S$ from the original speed to get $\frac{14}{15}S$. ### Final Answer Therefore, the correct answer is **5 km/hr**.