1. **Problem statement:** Ravi starts for school at 8:20 a.m. traveling at 10 km/h and arrives 8 minutes late. Traveling at 16 km/h, he arrives 10 minutes early. We need to find the school start time. 2. **Define variables:** Let the distance to school be $d$ km and the scheduled time to reach school be $T$ hours after 8:20 a.m. 3. **Write equations for travel times:** - At 10 km/h, time taken is $\frac{d}{10}$ hours, and he is 8 minutes ($\frac{8}{60}$ hours) late, so: $$\frac{d}{10} = T + \frac{8}{60}$$ - At 16 km/h, time taken is $\frac{d}{16}$ hours, and he is 10 minutes ($\frac{10}{60}$ hours) early, so: $$\frac{d}{16} = T - \frac{10}{60}$$ 4. **Set up system of equations:** $$\frac{d}{10} = T + \frac{2}{15}$$ $$\frac{d}{16} = T - \frac{1}{6}$$ 5. **Express $d$ from both equations:** $$d = 10\left(T + \frac{2}{15}\right)$$ $$d = 16\left(T - \frac{1}{6}\right)$$ 6. **Equate and solve for $T$:** $$10\left(T + \frac{2}{15}\right) = 16\left(T - \frac{1}{6}\right)$$ $$10T + \frac{20}{15} = 16T - \frac{16}{6}$$ 7. **Simplify fractions:** $$10T + \frac{4}{3} = 16T - \frac{8}{3}$$ 8. **Bring terms to one side:** $$10T - 16T = - \frac{8}{3} - \frac{4}{3}$$ $$-6T = - \frac{12}{3}$$ $$-6T = -4$$ 9. **Divide both sides by -6:** $$T = \frac{\cancel{-4}}{\cancel{-6}} = \frac{4}{6} = \frac{2}{3}$$ 10. **Interpret $T$:** $T = \frac{2}{3}$ hours = 40 minutes after 8:20 a.m., so scheduled time to reach school is 9:00 a.m. **Final answer:** The school starts at **9:00 a.m.**