SOLUTION: jeff leaves house at 8:30am and avgs 5mph, jane leaves at 9:00am avging 8mph at what time will jane catch jeff?

Question 372747: jeff leaves house at 8:30am and avgs 5mph, jane leaves at 9:00am avging 8mph at what time will jane catch jeff?
Answer by Jk22(389) (Show Source): You can put this solution on YOUR website!
jeff leaves house at 8:30am and avgs 5mph, jane leaves at 9:00am avging 8mph at what time will jane catch jeff?
at 9:00am jeff is 2.5miles from house.

after n hours after 9am, jane walked 8n miles, and jeff is 2.5+5n miles from the house,

they meet when they are at the same distance of the house :

8n = 2.5 + 5n
3n = 2.5
n = 2.5/3 = 5/6 hours, hence 50minutes

so 50 minutes after 9:00am, jane meets jeff, at 20/3miles=6.66miles from the house.

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