A student walks from his house at a speed of $2 \frac{1}{2}$ km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house? (S.S.C., 2007)

### Concept & Time Difference When a person travels the same distance at two different speeds, the distance can be found using the formula: $$Distance = \frac{S_1 \times S_2}{S_1 \sim S_2} \times \text{Difference in Time}$$ ### Step-by-Step Solution * **Given:** Initial speed $S_1 = 2 \frac{1}{2} = 2.5$ km/hr. Increased speed $S_2 = 2.5 + 1 = 3.5$ km/hr. * **Time Difference:** Late by 6 mins, then early by 6 mins. Total difference = $6 + 6 = 12$ minutes = $\frac{12}{60} = \frac{1}{5}$ hours. * **Calculation:** * $Distance = \frac{2.5 \times 3.5}{3.5 - 2.5} \times \frac{1}{5}$ * $Distance = \frac{8.75}{1} \times \frac{1}{5} = 1.75$ km. * $1.75 \text{ km} = 1 \frac{3}{4} \text{ km}$. ### Exam Strategy & Shortcut Use the direct formula $D = \frac{v_1 v_2}{\Delta v} \Delta t$. It avoids setting up algebraic equations and gives the answer in one step. ### Common Pitfall A frequent error is subtracting the times (6 - 6 = 0) instead of adding them when one is late and the other is early. ### Final Answer Therefore, the correct answer is **$1 \frac{3}{4}$ km**.