Two automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobile that left city A was 10 km/hour less than the speed of the other automobile. If the first automobile had left city A 4 1/2 hours earlier than the other automobile left city B, then the two would have met 150 km away from B. Find the distance between A and B.

Answer:450 kmStep-by-step explanation:Let's say Va is the speed of the car from city A, Ta is the time it spent traveling, and Da is the distance it traveled.Similarly, Vb is the speed of the car from city B, Tb is the time it spent traveling, and Db is the distance it traveled.Given:Va = Vb - 10Ta₁ = Tb₁ = 5Ta₂ = Tb₂ + 4.5Db₂ = 150Find:D = Da₁ + Db₁ = Da₂ + Db₂Distance = rate × timeIn the first scenario:Da₁ = Va Ta₁Da₁ = (Vb - 10) (5)Da₁ = 5Vb - 50Db₁ = Vb Tb₁Db₁ = Vb (5)Db₁ = 5VbSo:D = Da₁ + Db₁D = 10Vb - 50In the second scenario:Da₂ = Va Ta₂Da₂ = (Vb - 10) (Tb₂ + 4.5)Da₂ = Vb Tb₂ + 4.5Vb - 10Tb₂ - 45Db₂ = Vb Tb₂150 = Vb Tb₂Substituting:Da₂ = 150 + 4.5Vb - 10Tb₂ - 45Da₂ = 105 + 4.5Vb - 10Tb₂Da₂ = 105 + 4.5Vb - 10 (150 / Vb)Da₂ = 105 + 4.5Vb - (1500 / Vb)So:D = Da₂ + Db₂D = 105 + 4.5Vb - (1500 / Vb) + 150D = 255 + 4.5Vb - (1500 / Vb)Setting this equal to the equation we found for D from the first scenario:10Vb - 50 = 255 + 4.5Vb - (1500 / Vb)5.5Vb - 305 = -1500 / Vb5.5Vb² - 305Vb = -15005.5Vb² - 305Vb + 1500 = 011Vb² - 610Vb + 3000 = 0(Vb - 50) (11Vb - 60) = 0Vb = 50, 5.45Since Vb > 10, Vb = 50 km/hr.So the distance between the cities is:D = 10Vb - 50D = 10(50) - 50D = 450 km