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 =
`(1)/(18)` miles

So, Roger will run in x minutes =
`(x)/(18)` miles

Jeff runs in 15 minues = 1 mile

So, Jeff will run in 1 minute =
`(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 =
`(x-1)/(15)`

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

`(x)/(18) =(x-1)/(15)`

Cross Multiplying

`15 x = 18x - 18`

`3 x = 18`

x = 6 minutes

So for calculating distance =
`speed * time`

Distance =
`(1)/(18) *6 = (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 :)