Question 304959
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

The first cyclist rides at 6 mph.
So in 3 hours he rides 18 miles

Let the distance where they meet be x+18

By the time second cyclist travels x+18 miles, first cyclist rides x miles

Time taken by fist cyclist = distance / rate = x/6 hours

time taken by second cyclist = x+18 / 10

The times are same in both the case

x/6 = x+18/10

6(x+18)=10x

6x+108= 10x

4x=108

x= 27 miles

Distance / rate = time

x+18 /10 is the time taken by second cyclist to cross the first cyclist

27+18 /10 = 45/10 = 4.5 hours