Respuesta :
Alright, lets get started.
Roger can run one mile in 18 minutes.
Jeff can run one mile in 15 minutes.
Suppose in x minutes, they both will catch up.
Roger runs in 18 minutes = 1 mile
So, Roger will run in 1 minute = [tex]\frac{1}{18}[/tex] miles
So, Roger will run in x minutes = [tex]\frac{x}{18}[/tex] miles
Jeff runs in 15 minues = 1 mile
So, Jeff will run in 1 minute = [tex]\frac{1}{15}[/tex] mile
As Jeff gives roger a 1 minute head start, means Jeff will have x-1 minute time to run
So, Jeff will run in (x-1) minute = [tex]\frac{x-1}{15}[/tex]
As both are cathing up means they both runs same distance, means
[tex]\frac{x}{18} =\frac{x-1}{15}[/tex]
Cross Multiplying
[tex]15 x = 18x - 18[/tex]
[tex]3 x = 18[/tex]
x = 6 minutes
So for calculating distance = [tex]speed * time[/tex]
Distance = [tex]\frac{1}{18} *6 = \frac{1}{3}[/tex] miles
It means it will take 1/3 or 0.33 miles before jeff catches up to roger. : Answer
Hope it will help :)
