Correct Answer: B Let the total distance be $D$ km. Let the scheduled time to cover the distance be $T$ hours. We know that $\text{Distance} = \text{Speed} \times \text{Time}$. Therefore, $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$. **Case 1: Traveling at 40 km/hr** When the person travels at 40 km/hr, he reaches 15 minutes late. First, convert minutes to hours: $15 \text{ minutes} = \frac{15}{60} \text{ hours} = \frac{1}{4} \text{ hours}$. So, the time taken in this case is $T + \frac{1}{4}$ hours. Using the distance formula, we can write the equation: $D = 40 \times \left(T + \frac{1}{4}\right)$ (Equation 1) **Case 2: Traveling at 50 km/hr** When the person travels at 50 km/hr, he reaches 20 minutes early. Convert minutes to hours: $20 \text{ minutes} = \frac{20}{60} \text{ hours} = \frac{1}{3} \text{ hours}$. So, the time taken in this case is $T - \frac{1}{3}$ hours. Using the distance formula, we can write the equation: $D = 50 \times \left(T - \frac{1}{3}\right)$ (Equation 2) Since the distance $D$ is the same in both cases, we can equate the expressions for $D$ from Equation 1 and Equation 2: $40 \left(T + \frac{1}{4}\right) = 50 \left(T - \frac{1}{3}\right)$ Expand both sides of the equation: $40T + 40 \times \frac{1}{4} = 50T - 50 \times \frac{1}{3}$ $40T + 10 = 50T - \frac{50}{3}$ Now, rearrange the terms to solve for $T$. Move terms involving $T$ to one side and constant terms to the other side: $10 + \frac{50}{3} = 50T - 40T$ Combine the constant terms on the left side by finding a common denominator: $rac{30}{3} + \frac{50}{3} = 10T$ $rac{80}{3} = 10T$ Divide by 10 to find the value of $T$: $T = \frac{80}{3 \times 10}$ $T = \frac{8}{3}$ hours. Now that we have the scheduled time $T$, we can substitute its value back into either Equation 1 or Equation 2 to find the distance $D$. Using Equation 1: $D = 40 \times \left(T + \frac{1}{4}\right)$ Substitute $T = \frac{8}{3}$: $D = 40 \times \left(\frac{8}{3} + \frac{1}{4}\right)$ To add the fractions inside the parenthesis, find a common denominator (LCM of 3 and 4 is 12): $D = 40 \times \left(\frac{8 \times 4}{3 \times 4} + \frac{1 \times 3}{4 \times 3}\right)$ $D = 40 \times \left(\frac{32}{12} + \frac{3}{12}\right)$ $D = 40 \times \left(\frac{35}{12}\right)$ Multiply and simplify the expression: $D = \frac{40 \times 35}{12}$ Divide both 40 and 12 by their greatest common divisor, which is 4: $D = \frac{10 \times 35}{3}$ $D = \frac{350}{3}$ km. To express this as a mixed fraction, divide 350 by 3: $350 \div 3 = 116$ with a remainder of $2$. So, $D = 116 \frac{2}{3}$ km. Alternatively, using Equation 2: $D = 50 \times \left(T - \frac{1}{3}\right)$ Substitute $T = \frac{8}{3}$: $D = 50 \times \left(\frac{8}{3} - \frac{1}{3}\right)$ $D = 50 \times \left(\frac{7}{3}\right)$ $D = \frac{350}{3}$ km. Both equations consistently yield the same distance. The distance covered is $116 \frac{2}{3}$ km. The final answer is $\boxed{B}$.