Correct Answer: C) $200$ km\n\nLet $D$ be the total distance between City A and City B (in km).\nLet $S$ be the usual speed of the car (in km/hr).\nLet $T$ be the usual time taken for the journey (in hours).\n\nThe fundamental relationship is $D = S \\times T$.\n\n**Step 1: Analyze the first condition.**\nIf the speed is increased by $25\\%$, the new speed $S' = S + 0.25S = 1.25S$.\nThe car reaches 24 minutes earlier. So, the new time $T' = T - 24 \\text{ minutes} = T - \\frac{24}{60} \\text{ hours} = T - \\frac{2}{5} \\text{ hours}$.\n\nUsing the distance formula: $D = S' \\times T'$.\nSubstitute $D = S \\times T$: \n$S \\times T = 1.25S \\times (T - \\frac{2}{5})$\nSince $S \\neq 0$, we can divide by $S$:\n$T = 1.25 \\times (T - \\frac{2}{5})$\n$T = \\frac{5}{4} \\times (T - \\frac{2}{5})$\nMultiply by 4:\n$4T = 5T - 5 \\times \\frac{2}{5}$\n$4T = 5T - 2$\n$T = 2$ hours.\n\nSo, the usual time taken for the journey is $2$ hours.\n\n**Step 2: Analyze the second condition.**\nThe car travels $100$ km at its usual speed $S$.\nThe remaining distance is $(D - 100)$ km.\nFor this remaining distance, the speed is reduced by $20\\%$. So, the new speed $S'' = S - 0.20S = 0.8S$.\n\nThe car arrives 15 minutes later than the usual time. This means the delay is $15$ minutes, which is $\\frac{15}{60} = \\frac{1}{4}$ hours.\n\nSince the first $100$ km were covered at the usual speed, the delay is entirely due to the reduced speed over the remaining $(D-100)$ km.\n\nUsual time to cover $(D-100)$ km would be $t_{usual\_rem} = \\frac{D-100}{S}$.\nActual time taken to cover $(D-100)$ km with reduced speed is $t_{actual\_rem} = \\frac{D-100}{0.8S}$.\n\nThe delay is $t_{actual\_rem} - t_{usual\_rem} = \\frac{1}{4}$ hours.\n\n$\\frac{D-100}{0.8S} - \\frac{D-100}{S} = \\frac{1}{4}$\n\nFactor out $(D-100)$:\n$(D-100) \\left( \\frac{1}{0.8S} - \\frac{1}{S} \\right) = \\frac{1}{4}$\n\n$(D-100) \\left( \\frac{1}{4S/5} - \\frac{1}{S} \\right) = \\frac{1}{4}$\n\n$(D-100) \\left( \\frac{5}{4S} - \\frac{4}{4S} \\right) = \\frac{1}{4}$\n\n$(D-100) \\left( \\frac{1}{4S} \\right) = \\frac{1}{4}$\n\nMultiply both sides by $4$: \n$\\frac{D-100}{S} = 1$\n$D - 100 = S$\n\n**Step 3: Calculate the total distance $D$.**\nFrom Step 1, we know the usual time $T = 2$ hours.\nUsing $D = S \\times T$, we have $D = S \\times 2$, or $S = \\frac{D}{2}$.\n\nSubstitute $S = \\frac{D}{2}$ into the equation from Step 2 ($D - 100 = S$):\n$D - 100 = \\frac{D}{2}$\n\nMultiply by 2:\n$2D - 200 = D$\n$D = 200$ km.\n\nThus, the total distance between City A and City B is $200$ km.\n\nThe final answer is $\\boxed{\\text{C}}$