1. **Problem statement:** Bus A leaves at 12:00 noon and Bus B leaves 15 minutes later (12:15 pm). Bus B catches up with Bus A after covering 990 km at 1:20 pm. We need to find the speeds of both buses. 2. **Define variables:** Let $v_A$ be the speed of Bus A (in km/h) and $v_B$ be the speed of Bus B (in km/h). 3. **Time calculations:** - Bus A travels from 12:00 to 1:20 pm, which is 1 hour 20 minutes = $\frac{4}{3}$ hours. - Bus B travels from 12:15 to 1:20 pm, which is 1 hour 5 minutes = $\frac{13}{12}$ hours. 4. **Distance formula:** Distance = Speed $\times$ Time. - Distance traveled by Bus A in $\frac{4}{3}$ hours is $v_A \times \frac{4}{3}$. - Distance traveled by Bus B in $\frac{13}{12}$ hours is $v_B \times \frac{13}{12}$. 5. **Given:** Bus B catches Bus A after 990 km, so both have traveled the same distance: $$v_A \times \frac{4}{3} = 990$$ $$v_B \times \frac{13}{12} = 990$$ 6. **Solve for speeds:** - For Bus A: $$v_A = \frac{990}{\frac{4}{3}} = 990 \times \frac{3}{4} = 742.5$$ km/h - For Bus B: $$v_B = \frac{990}{\frac{13}{12}} = 990 \times \frac{12}{13} = 913.85$$ km/h (approx) 7. **Interpretation:** Bus A travels at 742.5 km/h and Bus B travels at approximately 913.85 km/h. These speeds are unusually high for buses, indicating a hypothetical or simplified problem.