Question

Roger can run one mile in 9 minutes. Jeff can run one mile in 6 minutes. If Jeff gives Roger a 1 minute head start, how

long will it take before Jeff catches up to Roger? How far will each have run?
Not including the head start, it will take
-
■
--
minutes for Jeff to catch up to Roger.

Answer 1

Answer: 2 minutes

Each person runs 1/3 of a mile when Jeff catches up to Roger.

================================================

Step-by-step explanation

x = number of minutes that Jeff runs

x+1 = number of minutes Roger runs

Roger has the head start of 1 minute, so he has been running for 1 minute longer compared to Jeff.

Roger runs 1 mile in 9 minutes. His unit rate is 1/9 of a mile per minute.

Jeff's unit rate is 1/6 of a mile per minute.

Let's set up a table with what we have so far

`\begin{array}c \cline{1-4} & \text{Distance} & \text{rate} & \text{time}\\\cline{1-4}\text{Jeff} & d & 1/6 & \text{x}\\\cline{1-4}\text{Roger} & d & 1/9 & \text{x}+1\\\cline{1-4}\end{array}`

The distance equation for Jeff is d = (1/6)x

The distance equation for Roger is d = (1/9)(x+1)

note: distance = rate*time

Both runners travel the same distance when Jeff catches up to Roger, so both "d"s are the same value at this specific moment. Set the right hand sides equal to each other and solve for x.

(1/6)x = (1/9)(x+1)

18*(1/6)x = 18*(1/9)(x+1)

3x = 2(x+1)

3x = 2x+2

3x-2x = 2

x = 2

Jeff runs for 2 minutes when he catches up to Roger.

----------

Check:

Jeff runs for 2 minutes, at 1/6 of a mile per minute, so he runs 2*(1/6) = 2/6 = 1/3 of a mile.

Roger runs for 2+1 = 3 minutes (remember he gets the head start) at 1/9 of a mile per minute, so he has run 3*(1/9) = 3/9 = 1/3 of a mile as well.

Both men have run the same distance which confirms Jeff catches up to Roger at this point. The answer is confirmed.