1. ## challenging rate problem

Dear Sir,
I would apppreciate very much if someone can help me in the below problem.
Thanks.

Tom and John were jogging along a circular track surrounding a pond. The track measured 640 m. If they started from the same place and jogged in the same direction, Tom would take 16 minutes to catch up with John. If they jogged in the opposite direction, they would meet every 4 minutes. How long did Tom take to jog one round of the track?

2. ## Re: challenging rate problem

@kingman

My GUESS(i'm not completely sure):

Algebraically:
640meters; Distance or circumference of the track
4mintues; How much time passes when Tom is passed by John
640meters/4mintues = 120meters; Distance Tom travels when he is passed by John
640meters/120meters = 8laps; Number of 120m segments
8laps * 4mintues = 32minutes; How much time it takes Tom to complete the track

NOTE: Units are incorrect.

Graphing:

@All

What are the three ways to solve math problems? I forgot.

3. ## Re: challenging rate problem

Since Tom must jog 1 lap (640 m) more than John before catching up,
then the problem is same as Tom starting off 1 lap (640m) behind John:
Code:
(640)            John........@ a mpm.............>16 min

Tom.......................@ (160 - a) mpm.....................>16 min
Since they meet every 4 min if starting in different directions, then
combined speed is 640/4 = 160 mpm (meters per minutes):
then, let John's speed = a mpm: then Tom's speed = 160 - a mpm.

Using the basic speed = distance / time formula, then for Tom:
(16a + 640) / (160 - a) = 16
Solve to get a = 60 : so Tom's speed = 160 - 60 = 100 mpm.

So time for Tom to jog 1 lap = 640 / 100 = 6.4 min = 6 min 24 sec.

4. ## Re: challenging rate problem

Originally Posted by ArcherSam
What are the three ways to solve math problems? I forgot.
LSP: Listen, Study, Practice !

How did you get this: "640meters/4mintues = 120meters" ? 640 / 4 = 160 ; no coffee yet?

5. ## Re: challenging rate problem

Originally Posted by Wilmer
Since Tom must jog 1 lap (640 m) more than John before catching up,
then the problem is same as Tom starting off 1 lap (640m) behind John:
Code:
(640)            John........@ a mpm.............>16 min

Tom.......................@ (160 - a) mpm.....................>16 min
Since they meet every 4 min if starting in different directions, then
combined speed is 640/4 = 160 mpm (meters per minutes):
then, let John's speed = a mpm: then Tom's speed = 160 - a mpm.

Using the basic speed = distance / time formula, then for Tom:
(16a + 640) / (160 - a) = 16
Solve to get a = 60 : so Tom's speed = 160 - 60 = 100 mpm.

So time for Tom to jog 1 lap = 640 / 100 = 6.4 min = 6 min 40 sec.

thanks but is'nt 6.4 mins equals 6 mins and 24 secs.

6. ## Re: challenging rate problem

Can you explain how you reason out that Tom must jog 1 lap (640 m) more than John before catching up.
They started at the same point ( for eg point A) and I wonder whether they meet at same point A after 16 mins.
thanks

7. ## Re: challenging rate problem

Originally Posted by kingman
thanks but is'nt 6.4 mins equals 6 mins and 24 secs.

8. ## Re: challenging rate problem

Originally Posted by kingman
Can you explain how you reason out that Tom must jog 1 lap (640 m) more than John before catching up.
They started at the same point ( for eg point A) and I wonder whether they meet at same point A after 16 mins.
thanks
A little "difficult" to "see", I agree; here's proof:
16 min @ 100 = 1600 m - 2*640 = 320
16 min @ 60 = 960 - 640 = 320
So Tom catches up at point 320 m , having gone around 1 extra "640".

Anyhow, YOU draw a circle and simulate what's happening; you'll see!

9. ## Re: challenging rate problem

I agree that it's difficult to see if they're both moving. But if you use the standard trick of letting John stand still the whole time and Tom run at the rate of T-J, where T is Tom's speed and J is John's, then Tom will complete a lap at this new speed in 16 minutes, giving 16(T-J)=640.

You also know that if you combine their speeds, they do a lap in 4 minutes, so 4(T+J)=640.

10. ## Re: challenging rate problem

Originally Posted by LoblawsLawBlog
But if you use the standard trick...
Agree LLB; however, I feel it is better for someone to first understand...then use shortcuts...