The correct answer is **B) 2 hours 40 minutes**. Here's a detailed step-by-step solution: **Step 1: Define variables** Let $D$ be the total distance covered by the bus (in km). Let $S$ be the usual speed of the bus (in km/hr). Let $T$ be the usual time taken by the bus (in hours). From the fundamental relationship of Time, Speed, and Distance, we have: $D = S \times T$ (Equation 1) **Step 2: Formulate equations from the given conditions** **Condition 1:** If the bus moves $10 \text{ km/hr}$ faster, it takes $30$ minutes less. * New speed: $S + 10$ km/hr * New time: $T - 30 \text{ minutes} = T - 0.5 \text{ hours}$ * Distance remains the same: $D = (S + 10)(T - 0.5)$ (Equation 2) Substitute Equation 1 into Equation 2: $ST = (S + 10)(T - 0.5)$ $ST = ST - 0.5S + 10T - 5$ $0 = -0.5S + 10T - 5$ Multiply by 2 to clear the decimal: $0 = -S + 20T - 10$ $S - 20T = -10$ (Equation 3) **Condition 2:** If the bus moves $10 \text{ km/hr}$ slower, it takes $45$ minutes more. * New speed: $S - 10$ km/hr * New time: $T + 45 \text{ minutes} = T + 0.75 \text{ hours}$ * Distance remains the same: $D = (S - 10)(T + 0.75)$ (Equation 4) Substitute Equation 1 into Equation 4: $ST = (S - 10)(T + 0.75)$ $ST = ST + 0.75S - 10T - 7.5$ $0 = 0.75S - 10T - 7.5$ Multiply by 4 to clear the decimal: $0 = 3S - 40T - 30$ $3S - 40T = 30$ (Equation 5) **Step 3: Solve the system of linear equations to find $S$ and $T$** We have two equations with two variables: 1. $S - 20T = -10$ 2. $3S - 40T = 30$ From Equation 3, we can express $S$ in terms of $T$: $S = 20T - 10$. Substitute this expression for $S$ into Equation 5: $3(20T - 10) - 40T = 30$ $60T - 30 - 40T = 30$ $20T - 30 = 30$ $20T = 60$ $T = 3 \text{ hours}$ Now, substitute the value of $T$ back into Equation 3 to find $S$: $S - 20(3) = -10$ $S - 60 = -10$ $S = 50 \text{ km/hr}$ Finally, calculate the total distance $D$ using Equation 1: $D = S \times T = 50 \text{ km/hr} \times 3 \text{ hours} = 150 \text{ km}$ **Step 4: Calculate the total time for the new journey scenario** The bus travels for the first $50 \text{ km}$ at its usual speed and then increases its speed by $10 \text{ km/hr}$ for the remaining journey. * **Part 1 of journey:** Distance = $50 \text{ km}$ Speed = Usual speed ($S$) = $50 \text{ km/hr}$ Time taken $t_1 = \frac{50 \text{ km}}{50 \text{ km/hr}} = 1 \text{ hour}$ * **Part 2 of journey:** Remaining distance = Total distance - Distance in Part 1 = $150 \text{ km} - 50 \text{ km} = 100 \text{ km}$ Speed = Usual speed + $10 \text{ km/hr} = 50 + 10 = 60 \text{ km/hr}$ Time taken $t_2 = \frac{100 \text{ km}}{60 \text{ km/hr}} = \frac{10}{6} = \frac{5}{3} \text{ hours}$ **Step 5: Calculate the total time** Total time = $t_1 + t_2 = 1 \text{ hour} + \frac{5}{3} \text{ hours}$ Total time = $\frac{3}{3} + \frac{5}{3} = \frac{8}{3} \text{ hours}$ **Step 6: Convert total time to hours and minutes** $\frac{8}{3} \text{ hours} = 2 \frac{2}{3} \text{ hours}$ $2 \text{ hours}$ and $\frac{2}{3} \times 60 \text{ minutes} = 2 \text{ hours and } 40 \text{ minutes}$. Thus, the total time taken by the bus is 2 hours 40 minutes.