Word Problem

Question

Roger can run one mile in 8 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?

Answer 1

Let x = number of minutes traveled by Roger until Jeff catches up

     

Then, x-1 = number of minutes traveled by Jeff

 

Distance = (Rate)(Time)

 

Roger's rate = 1mi/8min = 1/8 mi/min

 

Jeff's Rate = 1mi/6min  = 1/6 mi/min

 

Jeff catches up to Roger when Distance of Jeff = Distance of Roger

 

      (1/6)(x-1) = (1/8)x

 

      Multiply by the LCD, 24 to get 4(x-1) = 3x

 

                                                  4x - 4 = 3x

 

                                                        x = 4 minutes

 

      Jeff catches up to Roger 4 minutes after Roger started to run (3 minutes after Jeff started)

 

      Each of them ran (1/8)(4) = 0.5 miles