Question

Roger can run one mile in 18 minutes. jeff can run one mile in 15 minutes. if jeff gives roger a 1 minute head start, how long will it take before jeff catches up to roger?

Answer 1

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 = $\frac{1}{18}$ miles

So, Roger will run in x minutes = $\frac{x}{18}$ miles

Jeff runs in 15 minues = 1 mile

So, Jeff will run in 1 minute = $\frac{1}{15}$ 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 = $\frac{x-1}{15}$

As both are cathing up means they both runs same distance, means

$\frac{x}{18} =\frac{x-1}{15}$

Cross Multiplying

$15 x = 18x - 18$

$3 x = 18$

x = 6 minutes

So for calculating distance = $speed * time$

Distance = $\frac{1}{18} *6 = \frac{1}{3}$ miles

It means it will take 1/3 or 0.33 miles before jeff catches up to roger.   :   Answer

Hope it will help :)