A student goes to school at the rate of $2\frac{1}{2}$ km/h and reaches 6 min late. If he travels at the speed of 3km/h he is 10 min early. What is the distance to the school ?

### Concept & Time Difference Strategy When a person travels the same distance at two different speeds, the difference in the time taken is equal to the total time difference between arriving late and arriving early. We can equate the time difference using the formula: $$Time = \frac{Distance}{Speed}$$ ### Step-by-Step Solution 1. Let the distance to the school be $D$ km. 2. The initial speed is $2\frac{1}{2}$ km/h = $2.5$ km/h. The time taken is $D / 2.5$ hours. 3. The new speed is $3$ km/h. The time taken is $D / 3$ hours. 4. Total time difference = 6 minutes (late) + 10 minutes (early) = 16 minutes. 5. Convert the time difference to hours: $16 / 60$ hours. 6. Set up the equation: $$\frac{D}{2.5} - \frac{D}{3} = \frac{16}{60}$$ 7. Solve for $D$: $$\frac{3D - 2.5D}{7.5} = \frac{16}{60}$$ $$\frac{0.5D}{7.5} = \frac{16}{60}$$ $$\frac{D}{15} = \frac{16}{60}$$ $$D = 15 \times \frac{16}{60} = \frac{16}{4} = 4 \text{ km}$$ ### Exam Strategy & Shortcut Use the direct formula for finding distance when time difference ($\Delta t$) is given: $$Distance = \frac{S_1 \times S_2}{S_1 - S_2} \times \Delta t$$ $$Distance = \frac{2.5 \times 3}{3 - 2.5} \times \frac{16}{60}$$ $$Distance = \frac{7.5}{0.5} \times \frac{16}{60} = 15 \times \frac{16}{60} = 4 \text{ km}$$ This bypasses algebraic setups entirely. ### Common Pitfall Forgetting to convert the time difference from minutes to hours before applying it in the equation alongside speeds given in km/h. ### Final Answer Therefore, the correct answer is **4 km**.