Two airplanes left the same airport and arrived at the same destination at the same time. The first airplane left at 8:00 a.m. and traveled at an average rate of 496 miles per hour. The second airplane left at 8:30 a.m. and traveled at an average rate of 558 miles per hour. Let x represent the number of hours that the first plane traveled. How many hours did it take the first plane to travel to the destination? Enter an equation that can be used to solve this problem

Question
Answer:
If d is the distance involved 
then for first plane  d = 496x

and for second plane its d = 558/(x - 0.5)  , so:-

496x = 558x - 279

62x =  279

x = 4.5  hours

The first plane took 4 and a half hours



solved
general 10 months ago 8137