Two cars leave town at the same time and heading in the same

Two cars leave town at the same time and heading in the same direction. The first car travels 45mph and the second car travels 53mph. In how many hours will the second car be 30 miles ahead of the first?

Solution

Let the distance covered by the slower car be x miles in t hrs
so the distance covered by the fast car will be x+30 in t hrs
so time taken by the first car will be = distance/speed = x/45
and time taken by the second car will be = (x=30)/53
But the time taken is same t hrs
so the equation will be
x/45 = (x+30)/53
=> 53x = 45x + 45*30
=> 8x = 45*30
=> x = 45*30/8
=> x = 168.75
So the time will be 168.75/45 = 3.75
So in 3.75 hrs the second car will be 30 miles ahead of first