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, 4.5 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.

Given:

The two cars met at t= 5 hours
Car A leaves at t=0
Car B leaves at t = 4.5 hours, so only traveled for only 0.5 hours.
Car B travels 150 kilometers
so d = distance from A to B
As a result, Car A travels d - 150 km

150 = (Vb) * (0.5) ---> time of travel which is 0.5 hours
divide both sides by 0.5, you'll get Vb = 300 km/h this is the speed of Car B

Va = Vb - 10km/hr
Va = 300 - 10km/hr
Va = 290 km/hr is the speed of Car A

so:
d - 150 = (Va) * t
d - 150 = 290 * 5
d - 150 = 1450
d = 1450 + 150
d = 1600 km

The distance between Car A and Car B is 1600 kilometers.

40%=2/550=1/230=3/1056=14/25