Question

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 1

Answer:

450 km

Step-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 - 10

Ta₁ = Tb₁ = 5

Ta₂ = Tb₂ + 4.5

Db₂ = 150

Find:

D = Da₁ + Db₁ = Da₂ + Db₂

Distance = rate × time

In the first scenario:

Da₁ = Va Ta₁

Da₁ = (Vb - 10) (5)

Da₁ = 5Vb - 50

Db₁ = Vb Tb₁

Db₁ = Vb (5)

Db₁ = 5Vb

So:

D = Da₁ + Db₁

D = 10Vb - 50

In the second scenario:

Da₂ = Va Ta₂

Da₂ = (Vb - 10) (Tb₂ + 4.5)

Da₂ = Vb Tb₂ + 4.5Vb - 10Tb₂ - 45

Db₂ = Vb Tb₂

150 = Vb Tb₂

Substituting:

Da₂ = 150 + 4.5Vb - 10Tb₂ - 45

Da₂ = 105 + 4.5Vb - 10Tb₂

Da₂ = 105 + 4.5Vb - 10 (150 / Vb)

Da₂ = 105 + 4.5Vb - (1500 / Vb)

So:

D = Da₂ + Db₂

D = 105 + 4.5Vb - (1500 / Vb) + 150

D = 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 / Vb

5.5Vb² - 305Vb = -1500

5.5Vb² - 305Vb + 1500 = 0

11Vb² - 610Vb + 3000 = 0

(Vb - 50) (11Vb - 60) = 0

Vb = 50, 5.45

Since Vb > 10, Vb = 50 km/hr.

So the distance between the cities is:

D = 10Vb - 50

D = 10(50) - 50

D = 450 km