Correct Option: B Let $D$ be the total distance between City A and City B. Let $S$ be the original (scheduled) speed of the car. Let $T$ be the original (scheduled) time taken for the journey. Thus, $D = S \times T$. The car's speed is reduced by 20%, which means the reduced speed is $S - 0.20S = 0.8S$. Let's analyze the extra time incurred due to the reduced speed for a certain distance. If a distance $x$ is covered at the original speed $S$, the time taken is $\frac{x}{S}$. If it's covered at the reduced speed $0.8S$, the time taken is $\frac{x}{0.8S}$. The extra time for distance $x$ due to speed reduction is: $\text{Extra Time} = \frac{x}{0.8S} - \frac{x}{S} = \frac{5x}{4S} - \frac{4x}{4S} = \frac{x}{4S}$ This extra time is the delay observed in the problem statements. **Scenario 1:** Speed reduced after covering 100 km. The car travels 100 km at its original speed $S$. The remaining distance is $(D - 100)$ km. This remaining distance is covered at the reduced speed $0.8S$. According to the problem, the car reaches City B 45 minutes late. This entire 45-minute delay is caused by the extra time taken to cover the distance $(D - 100)$ km at the reduced speed. So, using our formula for extra time: $\frac{D - 100}{4S} = 45 \text{ minutes}$ Converting minutes to hours: $\frac{D - 100}{4S} = \frac{45}{60} \text{ hours} = \frac{3}{4} \text{ hours}$ This gives us our first equation: $4(D - 100) = 3(4S) \implies D - 100 = 3S \quad (Equation \ 1)$ **Scenario 2:** Speed reduced after covering 150 km. The car travels 150 km at its original speed $S$. The remaining distance is $(D - 150)$ km. This remaining distance is covered at the reduced speed $0.8S$. In this scenario, the car is 30 minutes late. This delay is due to the extra time taken to cover the distance $(D - 150)$ km at the reduced speed. Using the same formula for extra time: $\frac{D - 150}{4S} = 30 \text{ minutes}$ Converting minutes to hours: $\frac{D - 150}{4S} = \frac{30}{60} \text{ hours} = \frac{1}{2} \text{ hours}$ This gives us our second equation: $2(D - 150) = 1(4S) \implies D - 150 = 2S \quad (Equation \ 2)$ Now we have a system of two linear equations with two variables ($D$ and $S$): 1) $D - 100 = 3S$ 2) $D - 150 = 2S$ Subtract Equation 2 from Equation 1: $(D - 100) - (D - 150) = 3S - 2S$ $D - 100 - D + 150 = S$ $50 = S$ So, the original speed of the car, $S = 50$ km/hr. Now, substitute the value of $S$ into Equation 1 to find $D$: $D - 100 = 3(50)$ $D - 100 = 150$ $D = 250$ km The total distance between City A and City B is 250 km. The question asks for the scheduled (original) time for the car to travel from City A to City B, which is $T$. $T = \frac{D}{S} = \frac{250 \text{ km}}{50 \text{ km/hr}} = 5 \text{ hours}$. Thus, the scheduled time for the journey is 5 hours. The final answer is $\boxed{\text{5 hours}}$.