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Question

I have a difficult word problem I am confused by. Geoff walks the perimeter of the school in 15 minutes. Amia can walk the perimeter in 12 minutes. If they both start at the same place, how long will it take Amia to lap Geoff?

Answer 1

Geoff walks the perimeter in 15 minutes  ==>  

in 1 minute Geoff walks 1/15 of the perimeter  ==>  

in T minutes Geoff walks T*(1/15) = T/15 of the perimeter  

 

Amia walks the perimeter in 12 minutes  ==>  

in T minutes Amia walks T*(1/12) = T/12 of the perimeter  

 

Amia will lap Geoff when the distance she has walked (T/12 of the perimeter) is equal to one full lap more than Geoff has walked (T/15 of the perimeter plus 1), that is when  

 

T/12 = (T/15) + 1  

 

Multiply both sides by 60 to clear the fractions and get 

 

5*T = 4*T + 60

T = 60  

 

Amia will lap Geoff in 60 minutes  

 

Check:  After 60 minutes Amia has walked 60/12 = 5 laps

            After 60 minutes Geoff has walked 60/15 = 4 laps  

Answer 2

Say, the perimeter is x Mt

Speed of Geoff =x/15. Mt/min

Speed of Amia  =x/12. Mt/min

 

Amie is faster (walks more distance in one minute)

 

Amie walks x/12 meters in one minute

                  1 meter.        in 12/x minutes

                 x/15 meters.    in (12/x)(x/15)=12/15 minutes

                        = 0.8 minutes

So, after 0.8 minutes Amie  laps Geoff  

 

 

Answer 3

It may be best to use the perimeter to parametrize distance as its quantitiy is not given. Then Geoff's speed is P/15 and Amia's P/12 (Perimeters/min) and Amia is P(1/12 - 1/15)P = P/30  P/min) faster than Geoff. I terms of distance Amia is lead on Geoff is increasing by 1/30P*t with t in minutes. Amina will lap Geoff when the distance between becomes P. Then P/30*t=P or she will lap him in 30 minutes.