Question

Jack and Jill regularly run around an 800-meter track, but Jack runs 2 m/s faster. If they start from the same point on this track, and run in the same direction, Jack will eventually pass Jill in 6 minutes and 40 seconds. What is Jill's speed?

Answer 1

let the number of laps needed by Jack be n
then the number of laps Jill ran is n-1

Jack's distance = 800n
Jill's distance = 800(n-1)= 800n - 800

Jill's speed ---- x m/s
jack's speed --- x+2 m/s

time for the passing = 6 min, 40 s = 400s

Jill: 400x = 800n - 800
Jack: 400(x+2) = 800n
subtract them:
400(x+2) - 400x = 800
0 = 0

There is an infinite number of solutions, within a person's running speed.
e.g. let the number of laps completed by Jack be 10
then the number of laps completed by Jill is 9
distance covered by Jack = 8000 m
distance covered by Jill = 7200 m
Jacks speed = 8000/400 = 20 m/s (rather fast)
Jill's speed - 7200/400 = 18 m/s, which is 2 m/s slower

e.g. let the number of laps completed by Jack be 15
then the number of laps completed by Jill is 14
distance covered by Jack = 12000 m
distance covered by Jill = 11200 m
Jacks speed = 12000/400 = 30 m/s (really fast, outside possible range)
Jill's speed - 11200/400 = 28 m/s, which is 2 m/s slower

e.g. let the number of laps completed by Jack be 6
then the number of laps completed by Jill is 5
distance covered by Jack = 4800 m
distance covered by Jill = 4000 m
Jacks speed = 4800/400 = 12 m/s
Jill's speed - 4000/400 = 10 m/s, which is 2 m/s slower

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