Correct Option: C Let the distance be $D$ km. Let the original speed of the car be $S$ km/hr. Let the original time taken to cover the distance be $T$ hours. From the fundamental relationship of Time, Speed, and Distance, we have: $D = S \times T$ (Equation 1) **Condition 1: Speed increase** If the speed were 10 km/hr more, i.e., $(S+10)$ km/hr, the car would take 1 hour less, i.e., $(T-1)$ hours. So, we can write the equation: $D = (S+10)(T-1)$ Expanding this equation: $D = ST - S + 10T - 10$ Since $D = ST$ from Equation 1, we can substitute $ST$ for $D$: $ST = ST - S + 10T - 10$ Subtracting $ST$ from both sides: $0 = -S + 10T - 10$ Rearranging the terms to express $S$ in terms of $T$: $S = 10T - 10$ (Equation 2) **Condition 2: Speed decrease** If the speed were 5 km/hr less, i.e., $(S-5)$ km/hr, the car would take 1 hour more, i.e., $(T+1)$ hours. So, we can write the equation: $D = (S-5)(T+1)$ Expanding this equation: $D = ST + S - 5T - 5$ Again, substituting $ST$ for $D$ from Equation 1: $ST = ST + S - 5T - 5$ Subtracting $ST$ from both sides: $0 = S - 5T - 5$ Rearranging the terms to express $S$ in terms of $T$: $S = 5T + 5$ (Equation 3) **Solving for T and S** Now we have two expressions for $S$ (Equation 2 and Equation 3). We can equate them to solve for $T$: $10T - 10 = 5T + 5$ Subtract $5T$ from both sides: $5T - 10 = 5$ Add $10$ to both sides: $5T = 15$ $T = \frac{15}{5}$ $T = 3$ hours Now that we have the value of $T$, we can substitute it into either Equation 2 or Equation 3 to find $S$. Using Equation 3: $S = 5T + 5$ $S = 5(3) + 5$ $S = 15 + 5$ $S = 20$ km/hr **Calculating the Distance D** Finally, we can find the distance $D$ using Equation 1: $D = S \times T$ $D = 20 \text{ km/hr} \times 3 \text{ hours}$ $D = 60$ km The distance covered by the car is 60 km. The final answer is $\boxed{\text{60 km}}$